How powerful is $ACC^0$ circuit class in average case? We know that $NEXP$ is not in $ACC^0$ .

Does the result that $NEXP$ is not in $ACC^0$ also hold in average case? 

That is given a boolean function in $NEXP$ is it known that for every input length $n$ we cannot have $\frac1{n^c}$ ($c&gt;0$) fraction of possible computable in $ACC^0$? 

Tying up with communication complexity I think what I seek could be the following. Since $NEXP$ is not in $ACC^0$ is known could it be that any $NEXP$ complete function has superpolylogarithmic communication complexity in NOF model with superpolylogarithmic number of players? 

Then the query is could the randomized complexity still be just logarithmic. Is this possible?



Generally speaking what is the smallest class $\mathcal C$ that is known such that given a boolean function in $\mathcal C$ it is known that for every input length $n$ we cannot have $\frac1{n^c}$ ($c&gt;0$) fraction of possible inputs computable in $ACC^0$?
There seem to be typos in what you wrote; I take it you're asking: is there a function $f \in NEXP$ such that for all polynomials $p$ and infinitely many $n$, no ACC0 circuit of $p(n)$ size can compute $f_n$ on more than $1-1/p(n)$ inputs of length $n$?

As far as I know, this is open. But here is a possible path to doing it. We know that every $NEXP$-complete function $f$ does not have ACC0 circuits of polynomial size. Perhaps an arithmetization of some $f$ (representing $f$ naturally as a family of multilinear polynomials) over a large field of finite characteristic (e.g. characteristic two) would work. (Arithmetizations can typically give you average-case hardness "for free" ... but one would have to carry out the finite field arithmetic in ACC0, presumably.) Cite me if you figure it out. ;)