Degree $2$ nilpotent matrices with non-zero product and subset products not depending on the order Follow up to this answered question.
Let $n$ be sufficiently large positive integer. 
Let $S=\{M_i\}$ be a set of $n$ matrices with entries in either $\mathbb{C}$
or $\mathbb{Z}/m \mathbb{Z}$.
Sub-optimal statement, trying to keep the positive results.

Is it possible, (1) for $1 \le i \le n, M_i^2=0$ and
  (2) every product of the permutations of $\{S\}$ doesn't vanish: $\prod_{i} M_i \ne 0$ and (3) for every subset $t,t \subseteq \{S\}$, $\prod_{M \in t} M$ doesn't depend of the ordering of $t$, i.e. all orderings of $t$ have equal product.

(1) and (2) are possible as per answers in the linked question.
If it is possible, $M_i$ better be of as small dimension as possible.
Explicit construction for single $n$ as large as possible is of
interest to me.
 A: First of all, the condition "for every subset $t,t \subseteq \{S\}$, $\prod_{M \in t} M$ doesn't depend of the ordering of $t$, i.e. all orderings of $t$ have equal product" is equivalent to simply "matrices $M_i$ commute with each other". If this condition is satisfied for all 2-element $t$, then the matrices commute; and if matrices commute, then we can transform any ordering of a product into any other ordering by switching places of neighboring matrices.
Here is an example of such matrices: for $0\leq i\leq 2^n-1$, let $B_i$ be $2^n\times 2^n$ matrix with elements given by (here "&" denotes bitwise AND operator):
$$
(B_i)_{ab} = \begin{cases} 1 & \text{if $(a-1)\&i=0$ and $b=a+i$;}\\ 0 & \text{otherwise.}\end{cases}
$$
Then
$$
B_iB_j = B_jB_i = \begin{cases} B_{i+j} & \text{if}~~i\&j=0;\\ 0 & \text{if}~~ i\&j \neq 0.\end{cases}
$$
Set $M_i=B_{2^{i-1}}$. Then $M_i^2 = 0$ for all $i$, and $\prod\limits_{i=1}^n M_i = B_{2^n-1}$.
Now let's show that $2^n$ is the minimal possible size of the matrices. Set $I=\{1,\dots,n\}$. Let $e$ be a vector such that $\left(\prod\limits_{i=1}^n M_i\right)e\neq 0$, and for every $X\subseteq I$ set $e_X = \left(\prod\limits_{i\in X} M_i\right)e$ (in particular, $e_\varnothing = e$).
Let us prove that $\{e_X\}_{X\subseteq I}$ are linearly independent. Let some linear combination of them be zero:
$$
\sum\limits_{X\subseteq I} c_Xe_X = 0. ~~~~~~~~~\text{(*)}
$$
Assume that $c_X\neq 0$ for at least one $X$. Choose $Y\subseteq I$ such that $c_Y\neq 0$ and $c_X=0$ for all $X\subsetneq Y$. Multiply (*) by $\prod\limits_{i\in I\backslash Y} M_i$:
$$
0 = \sum\limits_{X\subseteq I} c_X\left(\prod\limits_{i\in I\backslash Y} M_i\right)e_X = \sum\limits_{X\subseteq I} c_X\left(\prod\limits_{i\in I\backslash Y} M_i\prod\limits_{j\in X} M_j\right)e.~~~~~~~~\text{(**)}
$$
By the choice of $Y$, if $X$ is a proper subset of $Y$, then $c_X=0$. And if $X$ is not a subset of $Y$, then there is $k\in (I\backslash Y)\cap X$, and
$$
\prod\limits_{i\in I\backslash Y} M_i\prod\limits_{j\in X} M_j = \left(\left(\prod\limits_{i\in (I\backslash Y)\backslash\{k\}} M_i\right)M_k\right)\left(M_k\left(\prod\limits_{j\in X\backslash\{k\}} M_j\right)\right) = 0
$$
since the product contains $M_k^2$. So we see that the only non-zero term in the rightmost sum of (**) is the one with $X=Y$:
$$
0 = c_Y\left(\prod\limits_{i\in I\backslash Y} M_i\prod\limits_{j\in Y} M_j\right)e = c_Y \left(\prod\limits_{i\in I} M_i\right) e.
$$
But by choice of $Y$ and $e$, $c_Y\neq 0$ and $\left(\prod\limits_{i\in I} M_i\right) e\neq 0$, so the right-hand side is non-zero. We have a contradiction.
So we have $2^n$ linearly independent vectors $\{e_X\}_{X\subseteq I}$, so the size of $M_i$ must be at least $2^n$.
