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.

For my answer, I assume that compositions are on non-negative numbers (allowing only positive numbers does not change my statements, qualitatively). First, observe that $p(N,m,n)=\binom{m}{n}_{N+1}$, where $\binom{m}{n}_{N+1}$ is a generalized, or extended, binomial coefficient (so there you have an exact formula, not only a lower bound). Since $\binom{m}{n}_A\ge \binom{m}{n}_B$ if and only if $A\ge B$, a (very weak) lower bound for $p(N,m,n)$ is $\binom{m}{n}$, the ordinary binomial coefficient.

However, much better lower bounds can be obtained by noting that $\binom{m}{n}_{N+1}$ arises as the distribution of the sum of $m$ iid random variables, uniformly distributed on $\{0,\ldots,N\}$. Then apply the central limit theorem (or a variant of the DeMoivre-Laplace theorem) to get nice bounds ...

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 $p(N,m,n)$.

However, instead of lower bounds, $\binom{m}{n}_{N+1}$ is also a nice (exact!) quantity.

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.