jmc

Note that when multiplying quadratics, terms add up similar to the equations of a system, so let$$\begin{array}{rl}p(x)& =(x^2+ax+c)(x^2+bx+d)\\ & =x^4+(a+b)x^3+(ab+c+d)x^2+(ad+bc)x+cd\\ & =x^4+15x^3+78x^2+160x+96\end{array}$$Factoring $p(x)$ with the Rational Root Theorem results in $(x+4)(x+4)(x+1)(x+6)$. By the Fundamental Theorem of Algebra, we know that $x+4,x+4,x+1,x+6$ are all the linear factors of the polynomial, so the quadratic factors can only be multiplied from these linear factors. There are only two possible distinct groupings (not counting rearrangements) -- $(x^2+8x+16)(x^2+7x+6)$ and $(x^2+5x+4)(x^2+10x+24)$. In the first case, $a^2+b^2+c^2+d^2=405$, and in the second case, $a^2+b^2+c^2+d^2=717$. The largest of the two options is $\boxed{717}$.