jmc

From the equation $a^2+b^2+c^2+d^2=4,$ $a^2\le 4,$ so $a\le 2,$ or $2-a\ge 0.$ Then $$(2-a)a^2\ge 0,$$so $a^3\le 2a^2.$ Similarly, $b^3\le 2b^2,$ $c^3\le 2c^2,$ and $d^3\le 2d^2.$ Adding all these inequalities, we get $$a^3+b^3+c^3+d^3\le 2(a^2+b^2+c^2+d^2)=8.$$Equality occurs when $a=2$ and $b=c=d=0,$ so the maximum value is $\boxed{8}.$