jmc

$$\begin{array}{rl}{x}^4-4x^3-4x^2+16x-8& =(x^4-4x^3+4x^2)-(8x^2-16x+8)\\ & =x^2(x-2{)}^2-8(x-1{)}^2\\ & =(x^2-2x{)}^2-(2\sqrt{2}x-2\sqrt{2}{)}^2\\ & =(x^2-(2+2\sqrt{2})x+2\sqrt{2})(x^2-(2-2\sqrt{2})x-2\sqrt{2}).\end{array}$$But noting that $(1+\sqrt{2}{)}^2=3+2\sqrt{2}$ and completing the square, $$\begin{array}{rl}{x}^2-(2+2\sqrt{2})x+2\sqrt{2}& =x^2-(2+2\sqrt{2})x+3+2\sqrt{2}-3\\ & =(x-(1+\sqrt{2}){)}^2-(\sqrt{3}{)}^2\\ & =(x-1-\sqrt{2}+\sqrt{3})(x-1-\sqrt{2}-\sqrt{3}).\end{array}$$Likewise, $$\begin{array}{r}{x}^2-(2-2\sqrt{2})x-2\sqrt{2}=(x-1+\sqrt{2}+\sqrt{3})(x-1+\sqrt{2}-\sqrt{3}),\end{array}$$so the roots of the quartic are $1\pm \sqrt{2}\pm \sqrt{3}$. Only one of these is negative, namely $1-\sqrt{2}-\sqrt{3}$, so the sum of the absolute values of the roots is $$(1+\sqrt{2}+\sqrt{3})+(1+\sqrt{2}-\sqrt{3})+(1-\sqrt{2}+\sqrt{3})-(1-\sqrt{2}-\sqrt{3})=\boxed{2+2\sqrt{2}+2\sqrt{3}}.$$