The solutions to the equation $(z+6)^8=256$ are connected in the complex plane to form a convex regular polygon, three of whose vertices are labeled $A,B,$ and $C$. What is the least possible area of $\triangle ABC?$

Here $R=256^{1/8}=2^{8/8}=2$. Using the same formula with $3$ consecutive octagon vertices:

Area $=\frac12R^2(\sqrt2-1)=\frac12(4)(\sqrt2-1)=\boxed{2\sqrt2-2}$.