A closed formula for $\det(\partial/\partial U)^p\prod_{i=1}^n\prod_{j=1}^p U_{i\sigma_j(i)}$

Question

This is a continuation of this question. Is there a simple formula for $$I(\sigma_1,\cdots,\sigma_p)=(-1)^\sigma\left(\left(\det\left(\frac{\partial}{\partial U_{ij}}\right)_{i,j=1}^n\right)^p \prod_{i=1}^n\prod_{j=1}^p U_{i\sigma_j(i)}\right)|_{U=0},$$ where all $\sigma_i\in S_n$? Equivalently $$I(\sigma_1,\cdots,\sigma_p)=\sum_{\pi\in C}\sum_{\tau_1,\tau_2\in R}(-1)^{\sigma\pi} [\tau_1\pi\tau_2=\sigma],$$ where $C$ and $R$ are the Young subgroups of $S_{pn}$ $$C=\mathrm{Sym}(1,\dots,n)\times\cdots\times\mathrm{Sym}((p-1)n+1,\dots,pn)\cong S_n^p$$ and $$R=\mathrm{Sym}(1,\cdots,(p-1)n+1)\times\cdots\times\mathrm{Sym}(n,\cdots,pn)\cong S_p^n$$ and $\sigma=\sigma_1\oplus\cdots\oplus\sigma_p\in C$.

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In the linked question, Carlo Beenakker guessed and Abdelmalek Abdesselam and David Speyer both proved that $I(\sigma_1,\sigma_2)=2^{\#\mathrm{cyc}(\sigma_1\sigma_2^{-1})}$ when $p=2$. This can also be expressed in the group algebra $\mathbb{C}[S_n]$ as $$\sum_{\sigma\in S_n} I(1,\sigma)\sigma=(2+J_1)\cdots(2+J_n),$$ where the $J_k$ are the Jucys-Murphy elements.

Note that $I$ is symmetric and $I(\sigma_1,\cdots,\sigma_p)=I(\rho\sigma_1,\cdots,\rho\sigma_p)$ for all $\rho$. When $(p,n)\ge(4,3)$ or $(3,5)$, $I$ can be negative. It appears that the prime factors of $I$ are at most $p$.

Here are the values of $I(\sigma_1,\sigma_2,1)/12$ when $p=3$ and $n=2$.

Here are the values of $I(\sigma_1,\sigma_2,1)/24$ when $p=3$ and $n=3$.

Here are the values of $I(\sigma_1,\sigma_2,1)/24$ when $p=3$ and $n=4$.

When $p=3$ and $n=5$, $I/24$ takes the values $-1,0,1,2,4,6,12,36,108,324$.

Answer 1

Perhaps for small values of $p$ one can find reasonable formulas but not for general $p$, I think. Take the cycle $\sigma_1=(12\cdots n)$ and then let $\sigma_j=(\sigma_1)^j$. Then, $I$ is, up to scale, $$ \int_{SU(n)} \prod_{i=1}^{n}\prod_{j=1}^{n} U_{ij}\ dU $$ which counts the difference between even and odd latin squares. It is an open problem, called the Alon-Tarsi conjecture to show that it is nonzero for $n$ even. See this article by Kumar and Landsberg.