Okay, I will note on EM that one part of it is straightforward to model. The size of the liquid air floods.
So air flows in at a known volumetric rate, liquifies, and flows outward again.
Inflow speed = mach 1 at some temperature, ~300m/s
Inflow surface area (bold for half sphere) = (300m radius)^2 * 4pi / 2 = 565,000 m^2
Inflow volume = 300m/s * 1,131,000 m^2 = 169,500,000 m^3/s
Volume reduction of air when it liquifies = air density / liquid air density = 1.29/870= 0.0015
Outflow volume of liquid air = 0.0015 * 169,500,000 m^3/s = 254,300 m^3/s
Right, so we have about 254k cubic meters of liquid air per second. Or, if spread over a square kilometer, a depth of 25 centimeters per second. So how far does the flood reach, from that? Well, that gets a little tricky. We could just guess a depth and then say the flood covers so-and-so kilometers per so-and-so, but lets be a little more rigorous.
What inhibits the liquid air flood? Well, it will touch warm ground and boil away, so it has to cool any ground it passes over. But the ground doesn't conduct heat all that well and the near-surface has only so much heat to contribute, so eventually it cools enough to pass over. So the liquid air finds a obstacle in the form of warm ground, and some of the air must boil away to cool that ground to pass over it, until eventually the ground is cooled to such a depth that it is no longer a relevant source of heat.
Energy to boil 1 kg of liquid air: ~200,000 joules/kg
Energy to boil 1 m^3 of liquid air: ~200,000 j/kg * 870kg/m^3 = 174,000,000 j/m^3
The amount of soil it has to cool is tricky to calculate. In part, it is a flood, so it will be churning topsoil while also simultaneously boiling on contact with that soil. And it could get complicated, with liquid air flowing into soil and then boiling after reaching non-frozen soil and that gas formation causes churning...
So lets look at permafrost. Permafrost is when the thermal mass and low conductivity of the ground averages out the yearly temperatures, so some portion of the soil below a certain depth is always frozen. You have a active zone at the top, which can freeze and unfreeze over the year, and below that it stays frozen. We can use this to eyeball how much work the liquid air has to do; if soil is deep enough to be frozen year-round, it's probably deep enough that it's thermal content is irrelevant to the much colder liquid air flood over the course of a few weeks. Permafrost depth varies depending on a few factors, but generally is a couple meters. Lets call it three.
Specific heat of soil: ~2 joules/gram/kelvin (significant variance)
Density of soil: ~1.4 g/cm^3
Specific heat of soil per cubic centimeter: 2 j/g/k * 1.4g/cm^3 = 2.8 j/k/cm^3
Specific heat of soil per cubic meter: 2.8j/k/cm^3 * (100cm)^3 = 2,800,000 j/k/m^3
Furthermore, soil contains water, which will freeze and release energy.
Water content of soil: varies significantly, assume ~10%
Water per cubic meter of soil: 10% * 1,400 kg/m^3 = 140kg/m^3
Heat of fusion of water: 333j/g, 333,000 j/kg
Heat of fusion of the water in one cubic meter of soil: 330,000 j/kg * 140kg/m^3 = 46,200,000 j/m^3
So, we need to create a temperature gradient from liquid air temperatures to normal soil temperature, across a depth of 3 meters of soil. This will be a linear gradient, so the average temperature drop will be one-half of the maximum. Furthermore, water will only freeze if the temperature drops by enough. The total temperature change is ~-220k from a room temperature of 20c, so we can assume that ~90% of the water along this temperature gradient freezes.
Energy to reduce the temperature of the relevant soil depth by the average temperature change: 110k * 2,800,000 j/k/m^3 * 3 m = 924,000,000 j/m^2
Energy to freeze the water in this region of soil: 90% * 46,200,000 j/m^3 * 3m = 138,600,000 j/m^2
Sum: 924,000,000 j/m^2 + 138,600,000 j/m^2 = 1,062,600,000 j/m^2
So we can expect soil to absorb around 1 gigajoule of energy per square meter from the liquid air flood. This means that each cubic meter will (eventually) boil away:
Energy to boil 1 m^3 of liquid air: 174,000,000 j/m^3
Energy required to fully cool a square meter of soil: 1,062,600,000 j/m^2
Depth of liquid air you have to pour onto soil to fully cool it: 1,062,600,000 j/m^2 / 174,000,000 j/m^3 = 6.1m
So we can reasonably assume any particular patch of land is thoroughly demolished, and no longer poses any particular obstacle to liquid air flowing over it, after it has boiled away ~6 meters of liquid air. Over the short term this number will be lower, but this is a good estimate for determining the long-term maximum size of the liquid air flood.
How much does this slow the flood? If it is required to pay the entire heat tax upfront, then...
Depth of liquid air you have to pour onto soil to fully cool it: 6.1m
Outflow volume of liquid air: 254,300 m^3/s
Square meters covered per second: 254,300 m^3/s / 6.1m = 41,700 m^2/s
Square kilometers covered per second: 41,700 m^2/s / (1000m)^2 = 0.0417/s
Of course, in practice, it does not have to pay the entire heat tax upfront. The more liquid air is flowing over any particular area of land, the less the heating from below is relevant. Near the EM Nuke itself, when first activated, the heating effect is likely small enough that fluid velocity is uninhibited. On the other hand, a greater flow of liquid air means greater disruption of the soil. If soil is dispersed throughout the liquid air by turbulent forces, then not only does it have no choice but to pay the heat tax immediately (as the surface area of contact is immense), but the total tax is increased as it has to cool the entirety of the displaced soil to the lowest temperature in addition to the undisturbed soil.
If, for instance, it physically displaces 3 meters of topsoil, the heat imparted onto the liquid air by this soil will be equivalent to that calculated above, and so the rate of expansion will be similar. As the actual amount of soil displaced will vary greatly by nuances of the flood that are difficult to calculate, it is also difficult to determine the actual rate of flooding near the start of the EM Nuke. Nonetheless, it may be slowed significantly, as the outflow is expected to be extremely turbulent and travel at potentially mach speeds, and so could potentially displace many meters of soil.
But the flood stops eventually; a infinite-duration EM Nuke would not just freeze the entire planet. This is because heat is getting added to any given patch of land... by the air, and sunlight. Ambient air is warm and will convey heat to the ground, and sunlight will constantly add heat to the system when it hits the ground and/or the cloud layers above the ground. So eventually the liquid air flood will be halted because it has covered such a area that the total energy received from sunlight is sufficient to boil it all and match the production rate.
Power from insolation per square meter, global average year-round: 340 j/s/m^2
Power from atmospheric heat transfer to the ground/liquid air flood: screaming in fluid dynamics = 0
Energy to boil 1 m^3 of liquid air: 174,000,000 j/m^3
Liquid air boiled per second per square meter of ground by sunlight: 340 j/s/m^2 / 174,000,000 j/m^3 = 0.0000020 m^3 boiled per m^2 of land per second (m^3/s/m^2, or m/s)
Per square kilometer: 0.0000020 m^3/s/m^2 * (1000m)^2 = 2m^3/s
Outflow volume of liquid air: 254,300 m^3/s
Square kilometers of land that will fully boil all of the outflowing liquid air: 254,300 m^3/s / 2m^3/s = 127,200 square kilometers
Radius of a circle with a area of 127,200 square kilometers: 201km
So, this puts a bound on the size of the liquid air flood. The average energy from sunlight is sufficient to cancel out the cooling effects of the EM AoE and boil all the liquid air across a circle with a diameter of 400 kilometers, and before that point the flooding will be slowed by the fact it is forced to extract heat from the soil it passes over. Heat transfer from the surrounding air contributes AAAAAAAAAAAAAAAAAAAAAAAAAAAAAAA, and wind effects are neglected.
So as you can see, it... uh. What was I talking about at the start of this?