What dB level corosponds to the lower 8 bits of a 24 bit pcm/wav file?

PCM samples in a wav file are coded in 2's complement.

For a N bits sample size, it means that the maximum positive sample value is 2^(N-1) - 1 and the minimum negative value is -2^(N-1).

For N = 24, it leads to a maximum positive value of 2^23 - 1. The maximum positive value of a sample composed of 8 lower bits set to 1 is 2^8 - 1.

So the dB FS value of such an 8 bit sample would be 20 * log10((2^8 - 1) / (2^23 - 1)), that is approximately -90.3 dB FS.

If you are measuring level with an RMS meter and your signal is a sinusoidal tone, the RMS meter should display about 3 dB less.

(Using 2^23instead of 2^23 - 1 as denominator has a negligible effect on the numerical result, although it is an interesting question to decide which one should be used).