What is the complexity for the universality problem for a NFA with all states accepting?
You problem is PSPACE-complete. I prove that it's PSPACE-hard by reducing universality of NFAs to universality of NFAs with all states accepting.



Let $A$ be a NFA. Add a new final state $q_\$$ to it and for every accepting state $q$, add a transition $q\overset{\$}{\to}q_\$$ where $\$$ is a fresh letter. For every letter $a\in \Sigma\sqcup \{\$\}$, also add a transition $q_\$\overset{a}{\to}q_\$$. And then make all states accepting. The new automaton accepts $L(A)\$ (\Sigma\sqcup \{\$\})^* \sqcup L'$ for some $L'\subseteq \Sigma^*$.

Now, add some automaton (with all states final) that recognizes the language $\Sigma ^*$ (i.e. the language of words that don't contain $\$$) in parallel. The new automaton will recognize $L(A)\$ (\Sigma\sqcup \{\$\})^* \sqcup (L' \cup \Sigma ^*)=L(A)\$ (\Sigma\sqcup \{\$\})^* \sqcup \Sigma ^*$.

Now, notice that we have $L(A)=\Sigma^*$ iff $L(A)\$ (\Sigma\sqcup \{\$\})^* \sqcup \Sigma ^*=(\Sigma\sqcup\{\$\})^*$. I.e. the process I described is a reduction from universality of a NFA to universality of a NFA with all states accepting. Since the reduction is polynomial, your problem is PSPACE-hard.