This is equivalent to finding the number of paths on a grid from $(0,0)$ to $(n,n)$ that never fall below the diagonal, and the answer is given by the Catalan numbers, which have the recurrence relation
$$C_{n+1}=\sum_{i=0}^{n}C_i\,C_{n-i}\quad\text{for }n\ge 0.$$
The Catalan numbers are one of those sequences that pop up again and again in combinatorics. They're worth knowing.
It would probably be instructive for you to convince yourself that your problem does in fact have the recurrence relationship above.
The general open-form solution could be of the form $\sum_{\begin{matrix}k_1,k_2,...,k_r\text{ such that }\\k_1+k_2+...+k_r=m\\k_1\in[0,min\{n_1,m\}]\\k_2\in[0,min\{n_2,m\}]\\...\\k_r\in[0,min\{n_r,m\}]\\k_i integers\end{matrix}}\left(\begin{matrix}m\\k_1k_2...k_r\end{matrix}\right)$.
Writing out each $n_i$ in a grid like format, it remains to find all possible combinations of values that sum to m given differing lengths of $n_1,n_2,...,n_r$, a seemingly challenging task.
$\begin{matrix}0 & 1 & 2 & 3 & \textbf{4} &... & n_1\\0&1&\textbf2&3&4&...&...&n_2\\...\\0&1&2&\textbf{$n_r$}\end{matrix}$
I have found it easy to begin small and consider the case with $\sum_{i=1}^rn_i=m$. In this case, You must pick all the things, and there is only one possibility. Then the answer is the multinomial coefficient with the corresponding values of $k_1=n_1,...,k_r=n_r$.
Now consider when $\sum_{i=1}^rn_i=m+1$. Looking at the hastily drawn image above, there is one degree of freedom that must be distributed across the r rows. There are r ways to do this. Now you must select all r combinations of $k_1,...,k_r$ such that they sum to m, plug that into the multinomial coefficient, and sum them up.
Consider when $\sum_{i=1}^rn_i=m+2$. Depending on r, the answer is different. $\begin{cases}r=1&r\\r\ge2&r+{r\choose2}\end{cases}$. When you've got only one type of item, you must pick all of them. But if you've got at least 2 types of things, you can distribute the 2 degrees of freedom to the same item (r possibilities) or distribute 1 each to 2 items (r choose 2). Note that this does not include actually finding the combinations of the $k_i$, this only tells you how many combinations there are.
Consider $\sum_{i=1}^rn_i=m+3$. Again, there are cases depending on r. $\begin{cases}r=1&r\\r=2&r+2{r\choose2}\\r\ge3&r+2{r\choose2}+{r\choose3}\end{cases}$. Notice the number of cases is how much smaller m is than n.
For $\sum_{i=1}^rn_i=m+4$, $\begin{cases}r=1&r\\r=2&r+2{r\choose2}+{r\choose2}\\r=3&r+2{r\choose2}+{r\choose2}+\left(\begin{matrix}3\\2,1\end{matrix}\right){r\choose3}\\r\ge4&r+2{r\choose2}+{r\choose2}+\left(\begin{matrix}3\\2,1\end{matrix}\right){r\choose3}+{r\choose4}\end{cases}$. Notice that a multinomial coefficient has appeared. I believe there will be at most one multinomial coefficient per "term" if you proceed this way, i.e. you won't have to worry about multinomial coefficients multiplying multinomial coefficients etc.
It has occurred to me that you can also start the other way, with $m=0$ and proceed upward.
Best Answer
Another way of looking at this is:
You have $n = n_1 + n_2 + \dots + n_r$ slots and need to fill them all.
You can fill in the $n_1$ items of type $1$ in $\binom{n}{n_1}$ ways.
The remaining $n-n_1$ slots can be filled with $n_2$ items in $\binom{n-n_1}{n_2}$ ways.
Continuing this way, the required number of ways is
$$ \binom{n}{n_1} \cdot \binom{n-n_1}{n_2} \cdots \binom{n-n_1-n_2-\dots-n_{r-2}}{n_{r-1}} \cdot 1 =$$
$$ \frac{n!}{n_1! (n-n_1)!} \cdot \frac{ (n-n_1)!}{(n-n_1-n_2)! n_2!} \cdots \frac{(n-n_1-n_2 - \dots -n_{r-2})!}{n_r! n_{r-1}!} = $$
$$ \frac{n!}{n_1! n_2! \dots n_r!}$$