Is it true that $$\sum_{r=0}^p \sum_{i=0}^r a_{n,p,r,i}=0$$ for all natural $n$ and all natural $p\ge2n$, where $$a_{n,p,r,i}:=\frac{(-1)^r (n+p-r-1)! (n p-i (r-i))}{i!(r-i)! (n-i)! (p-r+i)! (n-r+i)! (p-i)!}? $$
This is true if $n\in\{1,\dots,10\}$ and $p\in\{2n,\dots,2n+10\}$.
(Here it is assumed that $\dfrac1{j!}=0$ for $j\in\{-1,-2,\dots\}$.)
Subst $k = p - 2n \ge 0$ and $s = r - i$ to get the symmetric
$$\sum_{s \ge 0,i \ge 0} [s + i \le 2n + k] \frac{(-1)^{s+i} (3n+k-s-i-1)! (2n^2 + nk - is)}{i!(n-i)!(2n + k-i)! s!(n-s)!(2n + k-s)!}$$
But then we see from the $(n-i)!(n-s)!$ in the denominator that the bounds should actually be $0 \le i, s \le n$, and then we only get $x!$ for negative $x$ in the case $n = k = 0$, in the numerator:
$$\sum_{s=0}^n \sum_{i=0}^n \frac{(-1)^{s+i} (3n+k-s-i-1)! (2n^2 + nk - is)}{i!(n-i)!(2n + k-i)! s!(n-s)!(2n + k-s)!}$$
Sage gives Wilf-Zeilberger certificate for $s, i$ $$\frac{(2n^2 + kn - is + 3n + k - s - i)i}{(2n^2 + kn - is)(s + 1)}$$ so the inner sum is independent of $s$ and we have
$$n \sum_{i=0}^n \frac{(-1)^{s+i} (3n+k-s-i-1)! (2n^2 + nk - is)}{i!(n-i)!(2n + k-i)! s!(n-s)!(2n + k-s)!}$$
for an arbitrarily chosen $s$. We could just argue that by taking $s$ outside the support we get zero; alternatively, take $s=n$ and rearrange to
$$\frac{(-1)^n n}{n!(n-1)!(n+k)!} \sum_{i=0}^n (-1)^i \binom{n}{i}$$
The alternating binomial sum is well known to be zero.
