Determining if a word of specific length exists that is not accepted by a NFA It is known that the problem of determining if an NFA accepts every word is PSPACE-COMPLETE, meaning it is also NP-Hard, but is this weaker version of the problem still NP-hard?

Given an NFA and a natural number $n$, determine if there exists a word of length $n$ that is rejected by the NFA.

At first I thought that it is possible to prove this is NP-Hard by reducing the problem of checking if every word is accepted by an NFA to it, by checking every word length until hitting an upper bound. but according to Yuval Filmus in this question, the upper bound is exponential in the number of states of the NFA, so this reduction does not work.

Is there a valid reduction from a NP-Hard problem, or is this problem solvable in polynomial time?
Your problem is NP-hard, by reduction from 3SAT.

Let $\varphi$ be a 3SAT formula with $m$ clauses and on the $n$ variables $x_1,\dots,x_n$.  Construct a NFA over the alphabet $\Sigma=\{0,1\}$ as follows.  The $n$-bit input to the NFA is treated as an assignment to $x_1,\dots,x_n$.  The NFA first nondeterministically selects one of the $m$ clauses.  Then, it checks whether the input satisfies that clause, and if so, rejects; otherwise accepts.  Note that if the input $x$ satisfies $\varphi$, then the NFA rejects; otherwise, the NFA accepts.

Now there exists an input to the NFA of length $n$ that causes the NFA to reject, if and only if there is a satisfying assignment for $\varphi$.  That means that your problem is at least as hard as 3SAT.