The asymptotic version of this question raised by Bjorn Poonen has been studied by Khalid Bou-Rabee for general groups, not just free groups. That is, given G a residually finite group, for each g we can ask: how large is the smallest finite group F which detects g, meaning there exists f: G -> F so that f(g) is nontrivial? Now fix a word metric on G, and ask how the maximum of this "detection number" grows as you consider words of length at most n.
See "Quantifying residual finiteness" and "Asymptotic growth and least common multiples in groups" (with Ben McReynolds) for his results. For example, as long as G is linear, the growth function is polylog, meaning asymptotically less than (log n)k for some k, if and only if G is virtually nilpotent.
To answer your question, by considering congruence quotients of SL2Z, Bou-Rabee concludes that for every word of length n in the free group F2, there is a finite group of order O(n3) where the word is not an identity. The same bound can be obtained uniformly as follows. Ury Hadad gives a lower bound in "On the shortest identity in finite simple groups of Lie type" which implies that the shortest identity in PSL2(Fq) has length at least (q-1)/3. Since the size of PSL2(Fq) is order q3, this implies that every word of length at most n fails to be an identity in one single group PSL2(Fq) of order O(n3)!
I learned this argument from Martin Kassabov and Francesco Matucci's paper "Bounding the residual finiteness of free groups". In it, they use a nice analysis of finite groups with elements of "large order" to construct a word of length n in the free group F2 which is trivial in every finite group of order at most O(n2/3). This improved on the lower bound of n1/3 due to Bou-Rabee and Reynolds; I believe this is now the best lower bound known.