As an approximation, we assume $a, b, c, d$ are ordered to begin with (so we have to divide by 24 later) and add to 168 with a unique choice of signs; then, it suffices to count $e+f+g+h=168$ with each $e, f, g, h$ in $[-168,168]$ and then divide by 24 (we drop the condition that none of them can be zero because it shouldn't affect the answer that much). One way to do this is generating functions. We want the coefficient of $t^{168}$ in the generating function $\left(t^{-168}+t^{-167}+\ldots+t^{167}+t^{168}\right)^{4}=\left(t^{169}-t^{-168}\right)^{4} /(t-1)^{4}$. Clearing the negative powers, it suffices to find the coefficient of $t^{840}$ in $\left(t^{337}-1\right)^{4} /(t-1)^{4}=\left(1-4 t^{337}+6 t^{674}-\ldots\right) \frac{1}{(t-1)^{4}}$. To do this we expand the bottom as a power series in $t$: $\frac{1}{(t-1)^{4}}=\sum_{n \geq 0}\binom{n+3}{3} t^{n}$. It remains to calculate $\binom{840+3}{3}-4 \cdot\binom{840-337+3}{3}+6 \cdot\binom{840-674+3}{3}$. This is almost exactly equal to $\frac{1}{6}\left(843^{3}-4 \cdot 506^{3}+6 \cdot 169^{3}\right) \approx 1.83 \times 10^{7}$. Dividing by 24, we arrive at an estimation 762500. Even if we use a bad approximation $\frac{1}{6 \cdot 24}\left(850^{3}-4\right.$. $500^{3}+6 \cdot 150^{3}$) we get approximately 933000, which is fairly close to the answer.