Seemingly simple combinatorial problem Count all $n$-length strings of digits $0, 1,\dots, m$ that have an equal number of $0$'s and $1$'s. Is there a closed form expression?

This can be computed recursively. For $m\ge1$, let $f_m(n,d)$ be the number of $n$-letter strings over the alphabet $\{0,\dots,m\}$ containing exactly $d$ more $0$’s than $1$’s. Then $f$ satisfies the following, which provides a recursive definition:
$$\begin{align}
f_m(0,0)&=1\\
f_m(0,d)&=0\mbox{ if }d\ne0\\
f_m(n,d)&=f_m(n-1,d-1)+(m-1)f_m(n-1,d)+f_m(n-1,d+1)\\
\end{align}$$
The last of these equestions can be explained by considering how a string of length $n$ containing $d$ more $0$’s than $1$’s can be built by appending a digit to a string of length $n-1$. It can be built by appending a $0$ to a string among those counted by $f_m(n-1,d-1)$, by appending a digit other than a $0$ or $1$ (there are $m-1$ such digits) to a string counted by $f_m(n-1,d)$, or by appending a $1$ to a string counted by $f_m(n-1,d-1)$.
For $m=9$, this gives the following sequence of answers to the original question, beginning with $n=0$:
$1$, $8$, $66$, $560$, $4870$, $43248$, $390804$, $3582048$, $33213510$, $310850480$, $2931397756$.
This sequence does not appear (yet) in the Online Encyclopedia of Integer Sequences.
(These results agree with the sum Jack D’Aurizio gave in his answer.)
