jmc

Since $|a|=1,$ $a\overline{a}=|a{|}^2,$ so $\overline{a}=\frac{1}{a}.$ Similarly, $\overline{b}=\frac{1}{b},$ $\overline{c}=\frac{1}{c},$ and $\overline{d}=\frac{1}{d}.$

From the equation $a+b+c+d=0,$ $\overline{a}+\overline{b}+\overline{c}+\overline{d}=0,$ so $$\frac{1}{a}+\frac{1}{b}+\frac{1}{c}+\frac{1}{d}=0.$$This gives us $abc+abd+acd+bcd=0.$

Then by Vieta's formulas, $a,$ $b,$ $c,$ $d$ are roots of a polynomial of the form $$z^4+p_2{z}^2+p_0=0.$$If $z$ is a root of this polynomial, then so is $-z.$ This means $-a$ is equal to one of $b,$ $c,$ or $d,$ so $$(a+b)(a+c)(a+d)(b+c)(b+d)(c+d)=0.$$Therefore, the maximum value is $\boxed{0}.$