Indeed in the situation above

$$F^\nmid=\varinjlim_{\lbrace U_i\to U\rbrace} \ker \left(\prod_i F(U_i)\rightrightarrows \prod_{i,i'} F(U_i\times_U U_{i'})\right)=:H(\lbrace U_i\to U\rbrace, F)$$

is already zero. Fix a covering $\lbrace U_i\to U \rbrace$. By assumption we find $\lbrace V_{ij}\to U_i\rbrace$ such that $F(V_{ij})=0$ for all $i,j\in I\times J$. Also by the definition of a site $\lbrace V_{ij}\to U\rbrace$ is a covering.
The canonical refinement $\lbrace V_{ij}\to U\rbrace\to \lbrace U_i\to U\rbrace$ induces

$$
H(\lbrace U_i\to U\rbrace, F)\to H(\lbrace V_{ij}\to U\rbrace, F)=0
$$

where the right side is zero because the product of $F(V_{ij})$ is zero and so the kernel can't be anything else but zero.

Hence we have a system in which every object has a map to zero IN the system, and the colimit of such a system is zero as well.