Solution:
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 Vieta'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 $(-1, -1, -2015), (-1, -5, -403), (-1, -13, -155), (-1, -31, -65), (-5, -13, -31)$ 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 $8056 + 1440 = 9496$.