In the standard derivation for Sackur-Tetrode, the accounting for the indistinguishability of ideal gas molecules adds an extra factor of $N!$ in the partition function. This is usually approximated by Stirling's approximation. The entropy of an monoatomic gas without the large $N$ assumption is exactly (We are also assuming that the volume of the box is very large so that the energy spacing is very small, so we can replace the sum in the partition function with an gaussian integral.): $$S= Nk\left[\log(n_Q V)+\frac{3}{2}\right]-k \log N!$$ where $n_Q=\left(2\pi m k T/h\right)^{3/2}$ is an intensive quantity.
We can expand the Stirling series, $$S= Nk\left[\log(n_Q V)+\frac{3}{2}\right]-k \left(N \log N-N +\log\sqrt{2\pi N}+\mathcal O\left(\frac{1}{N}\right)\right)$$ The normal resolution of the Gibbs paradox is given by the truncation of the entropy at the leading order, $$S= Nk\left[\log(n_Q)+\log \frac V N+\frac{5}{2}\right]+ k \log\sqrt{2\pi N}+\mathcal O\left(\frac{1}{N}\right)$$ for which the term in the square brackets is extensive as one scales $N$ and $V$ simultaneously. It is said that this is how indistinguishability resolves the Gibbs paradox, so that entropy remains extensive. However, it is manifest that the subleading corrections do not scale properly.
What happens to the smaller terms at finite $N$? Does this mean Gibbs paradox isn't fully resolved, or we don't have extensive entropy? The more physical question might be, if we did an experiment with extremely dilute gases where $N$ is small, can we detect a non-extensivity? If not where does this calculation break down?