$(a+b+c)^2=0 \Longrightarrow \sum{a^2}=-2\sum{ab} \Longrightarrow \sum{a^4}+2\sum{a^2b^2} = 4\sum{a^2b^2}+8abc(a+b+c) = 4\sum{a^2b^2}\Longrightarrow 64 = \sum{a^2b^2}$, $\sum{a^4} = 128 \Longrightarrow (a^2+b^2+c^2)^2-2\sum{a^2b^2} = 128 \Longrightarrow a^2+b^2+c^2 = 16$, $(ab+bc+ac) = \frac{(a+b+c)^2-a^2-b^2-c^2}{2} = -8$. So only possible value is $\boxed{-8}$.