Since all the roots of $P(x)$ are integers, we can factor it as $P(x)=(x-r)(x-s)(x-t)$ for integers $r, s, t$. By Viete's formula, the product of the roots is $r s t=-2015$, so we need three integers to multiply to -2015. $P(x)$ cannot have two distinct positive roots $u, v$ since otherwise, $P(x)$ would be negative at least in some infinitesimal region $x<u$ or $x>v$, or $P(x)<0$ for $u<x<v$. Thus, in order to have two positive roots, we must have a double root. Since $2015=5 \times 13 \times 31$, the only positive double root is a perfect square factor of 2015, which is at $x=1$, giving us a possibility of $P(x)=(x-1)^{2}(x+2015)$. Now we can consider when $P(x)$ only has negative roots. The possible unordered triplets are $(-1,-1,-2015),(-1,-5,-(-1,-31,-65),(-5,-13,-31)$ which yield the polynomials $(x+1)^{2}(x+2015),(x+1)(x+5)(x+403),(x+1)(x+13)(x+155),(x+1)(x+31)(x+65),(x+5)(x+13)(x+31)$, respectively. Noticing that $P(-1)=0$ for four of these polynomials, we see that the nonzero values are $P(-1)=(-1-1)^{2}(2014),(5-1)(13-1)(31-1)$, which sum to $8056+1440=9496$.