I think that the answer is yes.  Here is a proof (hopefully blunder-free):

It is true for the presheaf $U \mapsto H^i(U,F\_1).$  In other words, if we fix $U$,
then for each element $h \in H^i(U,F\_1)$ and each geometric point $x$ of $U$,
there is an etale n.h. $V$ of $x$ such that $h\_{| V} = 0.$  Since $H^i(U,F\_1)$
is finite dimensional, there is a $V$ that works for the whole of $H^i(U,F\_1)$ at once.

I claim that then $H^i(U,F\_n)$ restricts to $0$ on $V$ as well.

To see this, consider the exact sequence $0 \to F\_n \to F\_{n+1} \to F\_1 \to 0.$
Applying $H^i(U,\text{--})$ to this yields a middle exact sequence
$H^i(U,F\_n) \to H^i(U,F\_{n+1}) \to H^i(U,F\_1).$   Applying $H^i(V,\text{--})$
yields a middle exact sequence
$H^i(V,F\_n)\to H^i(V,F\_{n+1}) \to H^i(V,F\_1).$  Restriction gives a map from the
first of these sequences to the second.  It is zero on the two outer terms, by induction
together with the case $n = 1$ proved above, and so is zero on the inner term.  

This shows that restricting from $U$ to $V$ kills $H^i(U,F_n)$ for all $n$, and hence
$H^i(U,F)$, as required.