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.
EDIT: As was noted in the comment below, this proof assumes that $F$ is ${\mathbb Z}_{\ell}$ -flat. Let me sketch an argument that hopefully handles the general case:
Put $F$ in a short exact sequence $0 \to F\_{tors} \to F \to F\_{fl} \to 0.$ The same kind of argument as above reduces us to checking $F\_{fl}$ and $F\_{tors}$ separately. The above proof handles the case of $F\_{fl}$, while $F\_{tors} = F\_{tors,n}$ for some large enough $n$, and so the projective limit collapses in this case and there is nothing to check.
(Note: I am assuming some basic kind of finiteness assumption on $F$ here, so that the above makes sense. Constructibility should be enough.)