Checking if all products of a set of matrices eventually equal zero
Your question is equivalent to whether $A_1, \dotsc, A_k$ generate a nilpotent algebra, which in turn is equivalent to each of the $A_i$ being nilpotent. Hence not only is it decidable, but in $\tilde O(n^{2 \omega})$ time where $\omega$ is the exponent of matrix multiplication. 

Let $\mathcal{A}$ be the associative algebra generated by the $A_i$: that is, take all linear combinations of the $A_i$ and all finite products thereof. $\mathcal{A}$ is called nilpotent if there is some $N$ such that every product of $N$ elements of $\mathcal{A}$ is zero.

First, let's see why your condition implies that $\mathcal{A}$ is nilpotent. This follows from Konig's Lemma (compactness): every string of length $n$ over the alphabet $\{1, \dotsc, k\}$ corresponds to a product of $A_1, \dotsc, A_k$ of length $n$ in an obvious manner. Consider the infinite $k$-ary rooted tree, whose nodes are naturally in bijective correspondence with strings over $\{1, \dotsc, k\}$. Consider the sub-tree $T$ consisting of those nodes where the corresponding product of the $A_i$ is nonzero. Konig's Lemma says that if $T$ is infinite, then it has an infinite path (exactly violating your property), hence $T$ is finite. We can then take $N$ to be the maximum length of any string in $T$. So your property implies that $\mathcal{A}$ is nilpotent.

The converse is also true, since every element of $\mathcal{A}$ is a linear combination of products of the $A_i$. 

Next, note that $\mathcal{A}$ is a subalgebra of $n \times n$ matrices, and hence is finite-dimensional.

Finally: a finite-dimensional associative algebra in characteristic zero has a basis of nilpotent elements (commuting or not - this is the part that contradicts Yuval's answer) iff it is nilpotent (see, e.g., here). 

Thus, to solve your problem, find a basis for the associative algebra generated by the $A_i$ (by the linear-algebra version of breadth-first search) and check that every matrix in the basis is nilpotent. The upper bound $\tilde O(n^{2\omega})$ comes from solving a system of linear equations in $n^2$ variables in the breadth-first search. As $\dim \mathcal{A} \leq n^2$ the BFS can't last very long, and because these are $n \times n$ matrices to check if a matrix $A$ is nilpotent one needs only check that $A^n = 0$.