This is question comes from a naive understanding of how these DSP algorithms work, but it seems to me that if you applied EQ to a certain frequency range that it is necessary to have additional bit depth available in order to maintain the incoming bit resolution across the range. For example, if you started with 16 bit data and used EQ with a difference of 18db between the equalised and unequalised sections of your response, you'd need to output an additional 3 bits per sample. If you did not, some part of your overall response would be only 13 bits resolution at full scale. It would not matter if you attenuate or boost, though this doesn't apply to crossover filters since the midband remains at full scale.

This is not really an issue of concern with 24 bit data, but it is possibly worth consideration if you're using a 16 bit sound card. Is this a valid assessment of the situation? How does Frequency Allocator handle this?

I believe FreqAllocator uses 64 bit processing internally. Therefore all calculations are completed (in 64bit precision) prior to the sound being sent to your card (16/24 bit).
Jan may correct me on this however.

You are correct. All of the processing including EQ and filtering is done with double precision - 64bit IIR math. It's one of the reasons Allocator gets praised for its sound quality.

The internal processing bit depth doesn't matter in this issue. If you begin with an input of 16 bits, apply EQ and then output 16 bits, you are losing some of the original bit depth due to the EQ. 18db of EQ amounts to 3 bits. Any part of the signal that is not boosted (or the attenuated part - depends how you view it) is only going to be represented by 13 bits.

16 bit in, 24 bit out, with 64 bit math regardless of the eq will preserve the original resolution. Once you attenuate more than 48dB you can start talking about loss of resolution, but then your analog stages with their noise floor are more of a dominant factor anyway.
Even if you use a 16bit converter but keep the signal normalized you don't really lose resolution. Your processed signal is not the same as your unprocessed signal anyway, so separating parts of it for comparison is only partially valid.

When applying EQ to flatten the output of a loudspeaker, if you normalise the data so that full scale is reached only on the loudest parts of the EQed signal, it stands to reason that the other, less loud sections of the EQed signal must be represented by fewer output bits. 18db is 3 bits. 16 - 3 = 13. What part of that reasoning is incorrect?

Admittedly, it isn't really an issue at all if you use a 24 bit soundcard and most do. And it is not a limitation specific to the Allocator - it is just mathematics.

I was thinking of using the EQ with an external, 16 bit DAC when this limitation occured to me.