2-dimensional ranking of multiple arrays
Here is a solution quite simple to implement, that only improves the $d$ factor.
For the increasing series $[1, 2, .., n]$, get the tuple list of the coordinates to increase in your output array. For $d = 2$ and $n = 4$, this is $H=[(1, 1), (1, 2), (1, 3), (1, 4), (2, 2), (2, 3), (2, 4), (3, 3), (3, 4), (4, 4)]$. Asymptotically (not exactly due to $(a, a)$ being always valid), there are $O(\frac{n^d}{d})$ terms in this list.
Initialize an output array $B$ of size $n^d$ to zeros.
Then, for each of the $m$ arrays $A$ and for each tuple $(a_0, a_1, ..., a_d)\in H$, increase $B[A[a_0], A[a_1], ..., A[a_d]]]$ by 1.
It achieves $O(\frac{m n^d}{d})$.
EDIT
For $d=3$, basically, both tuples $(0, 1, 2)$ and $(0, 2, 1)$ are in the coordinates list to increase. But if you want to use the knowledge that $B[a, b, c] = B[a, c, b]$, you have to keep only tuples non-decreasing after the 2nd integer. So here you leave $(0, 2, 1)$. Be careful to still keep $(a, b, b)$ tuples like $(0, 2, 2)$.
Then you will divide the number of terms in the list by about a factor $2(d-1)!$ for the general case).
At the very end, you have to add a $O(n^d)$ step to gather counts on $B[a, b, c]$ and $B[a, c, b]$. For every tuple $(a, b, c)$ with $b &lt; c$,

$S = B[a, b, c] + B[a, c, b]$
$B[a, b, c] = S$
$B[a, c, b] = S$

This runs in $O(\frac{m n^d}{d!})$.