Let $0 \leq m \leq n$ be integers. The group $S_n$ of permutations acts on the ring $\mathbb{Z}[X_1,\dots,X_n]$ by permuting the coordinates, with fixed subring $\mathbb{Z}[\sigma_1,\dots,\sigma_n]$, where the $\sigma_j$'s are the elementary symmetric functions.

Let $H$ be the stabilizer of $P = \prod_{j=m+1}^n X_j^m$, which is the subgroup of permutations leaving $[m+1,n]$ globally invariant, and consider the expansion $$ \prod_{\sigma \in S_n/H} (T - {}^{\sigma} P) = T^{\binom{n}{m}} + a_1 T^{\binom{n}{m} - 1} + a_2 T^{\binom{n}{m}-2} + \dots +a_{\binom{n}{m}}. $$ All $a_j$'s are fixed by $S_n$, hence can be written as $a_j = A_j(\sigma_1,\dots,\sigma_n)$, for a unique polynomial $A_j$.

Is it always true that the degree of $A_j$ is at most $jm$ ?

Note that if we give weight $j$ to $\sigma_j$, then $A_j$ is easily seen to have degree $jm(n-m)$, but here I'm asking for the degree of $A_j$ where we give weight $1$ to $\sigma_j$.

The answer is "yes" (and easy) when $m=0,1,n-1,n$. I also checked the case $m=2,n=4$, so that one can assume $n \geq 5$. I have strong reasons to believe, or at least to expect, that the result is true in general.