jmc

Since $(n-2)(n+4)(n+8)=0$ when $n=2,-4,$ or $-8$, we will consider the four cases $-11\le n<-8$, $-8<n<-4$, $-4<n<2$, and $2<n\le 11$ separately. If $n=2$, $n=-4$, or $n=-8$, then all three factors are 0. If $n>2$, then all three factors are positive. If $-4<n<2$, then $n-2$ is negative, while the other two factors are positive, so the product is negative. If $-8<n<-4$, then $n+8$ is positive, while the other two factors are negative, so the product is positive. If $n<-8$, then all three factors are negative, so the product is negative. In total, there are $\boxed{8}$ solutions: $-11,-10,-9,-3,-2,-1,0,1$.