As @MRC pointed stated, for NFAs the problem is also PSPACE-complete. However, I think the proof suggested by MRC does not cover the case you have in mind, when $k$ is non-constant, so I will give a complete proof here.
Lemma 1. With NFAs or regular expressions, the problem is PSPACE-complete.
Proof. First we give a proof for NFA variant. With DFAs the problem is known to be PSPACE-complete (shown originally by Kozen, I believe). The NFA variant cannot be easier (as the DFA variant is a restriction of the NFA variant), so the NFA variant is at least PSPACE-hard. It remains to verify that the NFA variant is in PSPACE.
It seems that the proof that the DFA variant is in PSPACE, as summarized here, adapts easily to the NFA variant. That proof uses Savitch's theorem, by which PSPACE = NPSPACE. So it suffices to describe a non-deterministic, polynomial-space verifier for the problem.
The input is a list of $n$ NFAs, each with at most $m$ states. The (non-deterministic, polynomial space) algorithm just non-deterministically guesses the string in the intersection, character by character, meanwhile simulating concurrently, in lockstep, each given NFA on the guessed string, character by character, keeping track of just one state per given NFA. If it ever reaches a step where all machines are in an accept state, it accepts. It requires only polynomial space, because at any point it only has to store $n$ states, one for each of the given NFAs.
This shows that the NFA variant is in PSPACE. The regular-expression variant reduces in polynomial time (and space) to the NFA variant, simply by converting each regular expression into an NFA by the standard polynomial-time algorithm. So the regular-expression variant is also in PSPACE. $~~~\Box$
By the way, the approach of explicitly constructing the NFA for the intersection of the given NFAs (using the standard Cartesian product construction) fails because that NFA can have as many as $m^n$ states, so in general has size exponentially large in the input size.