SNR of a Quantized Signal using FFT

It might be more practical to rephrase above post from fft perspective:

SNR = 6.02 * Bits + 1.76 + 10 Log ( M / 2 )

M = fft points

In your case SNR will be:

6.02*16 + 1.76 + 10log(8192/2) = 134.2dB

This level is what you will see.

However normalised SNR = above value minus effect of fft resolution. Thus normalised SNR becomes as you mentioned about theoretical floor.

normalised SNR = SNR - 10log(M/2)

Other factors to be considered:

upsampling will also increase SNR due to spreading. FFT window is another factor that may mask the fft noise floor.