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We give a deterministic algorithm for counting the number of satisfying assignments of any AC^0[oplus] circuit C of size s and depth d over n variables in time 2^(n-f(n,s,d)), where f(n,s,d) = n/O(log(s))^(d-1), whenever s = 2^o(n^(1/d)). As a consequence, we get that for each d, there is a language in E^{NP} that does not have AC^0[oplus] circuits of size 2^o(n^(1/(d+1))). This is the first lower bound in E^{NP} against AC^0[oplus] circuits that beats the lower bound of 2^Omega(n^(1/2(d-1))) due to Razborov and Smolensky for large d. Both our algorithm and our lower bounds extend to AC^0[p] circuits for any prime p.