Asymptotics of multinomial coefficients

Suppose that $k$ is a fixed natural number, $n\to\infty$, and \begin{equation} a_i=\frac nk+o(n^{2/3}) \end{equation} for each $i=1,\dots,k$. Let \begin{equation} h_i:=\frac kn\,a_i-1=o(n^{-1/3}). \end{equation} By Stirling's formula: \begin{equation} m!\sim\sqrt{2\pi m}(m/e)^m \end{equation} as $m\to\infty$,
\begin{equation} \binom{n}{a_1,\ldots,a_k} \sim(2\pi)^{1/2-k/2}\frac{n^{1/2}}{(n/k)^{k/2}}\frac{k^n}{e^u}, \end{equation} where \begin{equation} u:=\sum_{i=1}^k a_i\ln(1+h_i)=\frac nk\sum_{i=1}^k (1+h_i)\ln(1+h_i). \end{equation} Since $(1+h)\ln(1+h)=h+h^2/2+O(|h|^3)$ as $h\to0$, $\sum_{i=1}^k h_i=0$, and $h_i=o(n^{-1/3})$, we have \begin{equation} u=\frac n{2k}\,\sum_{i=1}^k h_i^2+o(1)=\frac k{2n}\,\sum_{i=1}^k(a_i-n/k)^2+o(1). \end{equation} Collecting the pieces, we get \begin{equation} \binom{n}{a_1,\ldots,a_k} \sim(2\pi n)^{1/2-k/2}k^{n+k/2}\exp\Big\{-\frac k{2n}\,\sum_{i=1}^k(a_i-n/k)^2\Big\}. \end{equation}

In particular, when $k=2$, we get the Wikipedia result quoted in the OP: \begin{equation} \binom na\sim\frac{2^n}{\sqrt{\pi n/2 }} e^{-2(a-n/2)^2/n} \end{equation} if $n\to\infty$ and $a=\frac n2+o(n^{2/3})$.

The analogous estimate in the range where your formula is valid (that is, $a_i=\frac nk+o(n^{1/2})$ - note Wikipedia claims the binomial case is valid for $a_i=\frac n2+o(n^{2/3})$, but I wasn't able to reproduce that estimate) is: $$ \binom{n}{a_1\ a_2\ \ldots\ a_k} \sim \frac{k^n}{(2\pi n)^{(k-1)/2}}\exp\left(-\frac kn\sum_{i=1}^k b_i^2\right), $$ where $b_i=a_i-\frac nk$, that is the difference of $a_i$ from its central value.

This can be obtained by using Stirling's formula, and estimates on quantities of the form $(1+\frac an)^n$.