Well, if $N\ge 1$, then I think that $\binom{m}{n}$ is a lower bound (if you allow $0$'s in the compositions). In fact, $p(N,m,n)=\binom{m}{n}_{N+1}$, where $\binom{m}{n}_{N+1}$ denotes a generalized, or extended, binomial coefficient, and it holds that $\binom{m}{n}_{A}\ge \binom{m}{n}_B$, for all $m$ and $n$, if and only if $A\ge B$. So, $\binom{m}{n}_2=\binom{m}{n}$ would be a (very weak, though) lower bound for $p(N,m,n)$. (If your compositions are on positive integers, the corresponding lower bound would be $\binom{m}{n-m}$)
However, instead of lower bounds, $\binom{m}{n}_{N+1}$ is also a nice (exact!) quantity.