it seems that the heat at / near the surface will determine the melt, not the heat 1,000 feet up.

It's a fair question. The short answer is that the air near the surface holds hardly any heat energy because it's a gas and so it contains hardly any mass per unit volume.

In more detail, these 3 points are relevant:

1) If we are just looking at the air temperature then we are ignoring water vapor, which can be important. (The dew point shows how much water vapor is in the air.) But now considering just dry air...

2) The temperature at 2 meters is not representative of the amount of heat in the air column as it is somewhat tied to the ~0 degrees temperature just below it.

3) The (dry) air in the meters just above the ice is carrying very little energy and so can melt only a negligible depth of ice. To get significant melt, a significant fraction of the heat in the air column has to be transported somehow into melting the ice (water vapor, infrared radiation, convection...)

To illustrate point 3, consider how much warm air would be required to melt, e.g., a 1 cm thickness of ice.

(We assume some quasi-static conditions where the air column above the ice causes the melt directly below it - with the caveat that this is not normally a very good assumption.)

To melt, e.g., a 1 cm depth of ice requires (334 J/g specific heat of melt) x 0.9 g/cm^3 = 300 J/cm^2.

So 300 J would need to be supplied by the air column above each square cm of ice.

But the specific heat of air is only 1 J/g.(degree C) and the mass of air in the entire column all the way up into space is only about 1000 g/cm^2 (i.e one atmosphere) What average loss of air temperature, dT_air, would be needed to supply the 300 J/cm^2 to melt a 1 cm thickness of ice?

Require:

dT_air[Celsius] x 1.0 J/g.[degree C] x 1000 g/cm^2 = 300 J/cm^2

=> dT_air = 0.3 degrees C.

So THE ENTIRE COLUMN OF AIR UP INTO SPACE would need to lose 0.3 degrees C to melt a 1 cm depth of ice. Obviously, the air just directly above the ice can't melt much ice at all.