jmc

By AM-GM, $$a+b+c+d\ge 4\sqrt[4]{abcd}=4\sqrt[4]{10!}\approx \mathrm{174.58.}$$Since $a,$ $b,$ $c,$ $d$ are all integers, $a+b+c+d\ge 175.$

Note that $a=40,$ $b=42,$ $c=45,$ and $d=48$ satisfy $abcd=10!,$ and $a+b+c+d=\boxed{175},$ so this is the minimum.