The analogous formula for the permutation characters of the symmetric group induced from Young subgroups is that $\mathrm{Ind}_{S_\lambda}^{S_n} 1_{S_\lambda} = \sum_{\mu \unrhd \lambda} K_{\mu\lambda} \chi^\mu$. For instance the case $\lambda = (1^n)$ is the regular representation.
It therefore appears that there is a typo and $\mu$ and $\lambda$ should be swapped in the formula in Stanley's review to give
$$\eta_\lambda(q) = \sum_{\mu \unrhd \lambda} K_{\mu\lambda}\chi^\mu(q).$$
Then taking $\lambda = (1^n)$ we get $\eta_{(1^n)}(q) = \sum_{\mu} K_{\mu(1^n)}(q)\chi^\mu(q) = \sum_{\mu} |\mathrm{SYT}(\mu)|\chi^\mu(q)$, corresponding to the fact that every unipotent character appears in the induction of the trivial character of the Borel. And taking $\lambda = (n)$ when the parabolic subgroup $P_{(n)}$ is the full general linear group, we get $\eta_{(n)}(q) = \chi^{(n)}(q) = 1_{\mathrm{GL}_n(\mathbb{F}_q)}$, also correct.