How can you shuffle in $O(n)$ time if you need $\Omega(n \log n)$ random bits?
Your argument appears to be perfectly valid (Fisher-Yates does indeed require $\log (n!)$ bits of randomness), the discrepancy comes in by making different assumptions about the complexity of the random number generation.

You're assuming generating a random number between $0$ and $n$ takes $O(\log n)$.

But, when saying that the Fisher-Yates shuffle is $O(n)$, one assumes generating a random number takes $O(1)$.

An integer index (which is used in the Fisher-Yates shuffle) is generally considered $O(1)$ space (even though it's technically $O(\log n)$) because you can index way more data with a simple 64-bit integer than can currently reasonably fit into the memory of any computer most of us have access to $^{1}$. And the space complexity per number is same as the complexity for the number of random bits that must be generated for that number.

1: To go beyond the number of elements a 64-bit integer can index, we would require $2^{64}$ = $18446744073709551616$ elements, and, assuming 1 bit (which is obviously very small) per element, this requires 2 exabytes.