jmc

Moving all the terms to the left-hand side, we have $$z^4+4i{z}^3-6z^2+(8-4i)z+(1+8i)=0.$$Seeing the coefficients $4$ and $6$ reminds us of the expansion for $(z+1{)}^4.$ To get terms such as $4i{z}^3$ which involve $i,$ we instead write $$(z+i{)}^4=z^4+4i{z}^3-6z^2-4iz+1.$$In view of this, the given equation is equivalent to $$(z+i{)}^4+8z+8i=0,$$or $$(z+i{)}^4=-8(z+i).$$Making the substitution $w=z+i,$ we have $$w^4=-8w.$$Because this substitution only translates the complex plane, the sum of the pairwise distances does not change if we work with this equation instead of the equation for $z.$ This equation implies that either $w=0$ or $$w^3=-8.$$Every solution to $w^3=-8$ has magnitude $2$, because taking magnitudes of both sides gives $|w^3|=|w{|}^3=8.$ Furthermore, if $w^3=-8,$ then $w^6=64,$ so $w$ is two times a number that is a $6^{\text{th}}$ root of unity that is not a $3^{\text{rd}}$ root of unity. These complex numbers have arguments $\frac{\pi }{3},$ $\pi ,$ and $\frac{5\pi }{3}$ in the complex plane, so they form an equilateral triangle: This equilateral triangle has side length $2\sqrt{3},$ so its perimeter is $6\sqrt{3}.$ Together with the distances of $2$ from each vertex to the origin, we get the answer, $6\sqrt{3}+2(3)=\boxed{6\sqrt{3}+6}.$