Binomial coefficients have a well known asymptotics (<https://en.wikipedia.org/wiki/Binomial_coefficient#Bounds_and_asymptotic_formulas>) given by $$\binom nk\sim\binom{n}{\frac{n}{2}} e^{-d^2/(2n)} \sim \frac{2^n}{\sqrt{\frac{1}{2}n \pi }} e^{-d^2/(2n)}.$$


Is there corresponding asymptotics for multinomials (I am more interested in what happens analogously in the denominator)?



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 $i$; here in what follows, $i\in\{1,\dots,k\}$.
Let
\begin{equation\*}
 h\_i:=\frac kn\,a\_i-1=o(n^{-1/3}), \tag{1}
\end{equation\*}
so that $h\_i\to0$ and
\begin{equation}
 a\_i=\frac nk\,(1+h\_i). \tag{2} 
\end{equation}
By Stirling's formula,
\begin{equation\*}
 n!\sim(2\pi)^{1/2} n^{1/2}\Big(\frac ne\Big)^n
\end{equation\*}
and, by (2),
\begin{equation\*}
\begin{aligned}
 a\_i!&\sim(2\pi)^{1/2} \Big(\frac nk\Big)^{1/2}(1+h\_i)^{1/2}\Big(\frac n{ke}(1+h\_i)\Big)^{a\_i} \\ 
 &\sim(2\pi)^{1/2} \Big(\frac nk\Big)^{1/2}\Big(\frac n{ke}(1+h\_i)\Big)^{a\_i}. 
\end{aligned}
\end{equation\*}
Therefore and because $\sum\_{i=1}^k a\_i=n$,
\begin{equation\*}
 \prod\_{i=1}^k a\_i!\sim(2\pi)^{k/2} \Big(\frac nk\Big)^{k/2}\Big(\frac n{ke}\Big)^n
 \prod\_{i=1}^k(1+h\_i)^{a\_i}, 
\end{equation\*}
which implies
\begin{equation\*}
 \binom{n}{a\_1,\ldots,a\_k}=\frac{n!}{\prod\_{i=1}^k a\_i!}
 \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\*}
by (2).
Note also that (i) $(1+h)\ln(1+h)=h+h^2/2+O(|h|^3)$ as $h\to0$, (ii) $\sum\_{i=1}^k h\_i=0$ (by the definition of $h\_i$ in (1) and the condition $\sum\_{i=1}^k a\_i=n$), and (iii) $h\_i=o(n^{-1/3})$. It follows that
\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\*}
Thus,  

\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 $a=\frac n2+o(n^{2/3})$.


