Almost none of the energy goes into the air. Air is barely heated by radiation, because it is transparent.
What happens is that the re-entry heating of the falling ejecta produces radiative heating, which directly heats the surface of the earth. And that gets cooked well past oven temperature.
The nice thing of energy conservation is that the details doesn't mater. So the energy of the radiative heating is less than the original energy of the meteorite that I estimated in 9.5E22 J. I guess it's much less, but it's a good upper bound.
The surface of the Earth is 5.1E14 m2, so it's 1.8E8 J/m2. I don't expect the fireworks to be synchronized so let's assume that they fall in an hour. That gets 50,000 W/m^2 For comparison, sunlight is 1,300 W/m^2 so it's almost like 40x sunlight. This bad, very bad.
I don't expect a 100% conversion of meteorite energy to fallout energy, but it's more dangerous that my initial calculation.
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I'm confused because the weight of the atmosphere is like the weight of 10m of water. So heating all the atmosphere is like heating a 10m layer of water. But actually water has a much bigger heat capacity, so the final temperature would be smaller. And dirt has an intermediate value. Also the radiation would not heat 10m of water or dirt, perhaps only the first meter.
So I expected that if my first calculation got a not dangerous temperature increase, I'd get in this second calculation a not dangerous result. So I guess I made a mistake in at least one of the calculations or my hand waving has a mistake.
What happens is that the re-entry heating of the falling ejecta produces radiative heating, which directly heats the surface of the earth. And that gets cooked well past oven temperature.