Is computing 2^n NP-complete problems EXPTIME or NEXPTIME complete?
I am not 100% sure that I understand the question but here is an attempt where I reformulate your problem.

You have $A,B\subseteq \{0,1\}^*$ and know that $A$ is NP-complete. Furthermore you have a reduction $f$ such that $f(x)\in B$ iff $x\in A$ and such that $|f(x)|$ is in $\log(O(|x|)$. Does that make $B$ NEXPTIME-hard?

Well, since $A$ is NP-complete, I can translate any question $y$ of the the form "does a nondet. TM $M$ halts in time $n$?" into an equivalent question $g(y)$ for $A$, using the reduction that establishes the NP-hardness of $A$. It is thus possible to translate an NEXPTIME-hard question $y$ of the form "does $M$ halts in time $2^n$?" into an equivalent $g'(y)$ for $A$. That reduction is very simple to describe but it produces instances of exponential size. If we could compute $f(g'(x))$ efficiently, we would have a reduction showing that $B$ is NEXPTIME-hard. This is possible in many cases --remember that $g'$ is very simple and that $|f(g'(x))|$ is in $O(|x|)$-- but this depends on knowing $f$ and how nicely it interacts with $g'$. 

So I'd say that in general you cannot say that $B$ is NEXPTIME-hard but in many specific cases you'll manage to show it is.

(Final note: the above is about lower bounds. You cannot provide upper bounds for the complexity of $B$ from the assumptions. They only talk about the part of $B$ that is in the image of $f$).