But... the OAT is lower as the pressure is lower. As you raise the pressure in the induction system, the temperature will rise again. Now, is the atmosphere lapse rate with pressure higher or lower than what you get as you compress the air in the induction system? That determines whether the manifold air is warmer or colder than it was at a lower altitude, for the same MAP. I'm not going to venture a guess either way, but I'm willing to guess that the temperature difference won't be enough to matter.
-------- Quick-and-dirty and possibly poorly thought through thought experiment below - do not use for operating nuclear power plants, and ignore unless willing to risk subjecting yourself to gross errors - corrections welcome -------
Let's forget those pesky non-ideal properties of the atmosphere for a bit and play with the ideal gas law.
pV=nRT
p = nRT/V
n = pV/RT
Density (n) and pressure go hand in hand, unless the temperature changes. However, the temperature does change. Using your figures to calculate the ratio of T at altitude over T at SLS,
6877 = 7385*R*T/V
while SLS gives
1 = 1*R*T/V, or R*T/V = 1.
At altitude, that'd mean R*T/V = 6877/7385, or .931
R constant and V constant gives
T ratio = .931 * (273+15) K = 268 K = -5 C
Just what my CR-3 says it should be, in the standard atmosphere, so the ideal gas law seems to be valid to use for the lapse rate. Should be equally valid, within reasonable limits, as you bring p up again through the compressor of the turbo/supercharger giving you the same T in the manifold regardless of altitude. Hence, we need to look at other reasons for the power increase with altitude with constant MAP - and back pressure is one.