解:$(1)$∵$\sqrt {16}<\sqrt {17}<\sqrt {25},$即$4<\sqrt {17}<5$
∴$1<\sqrt {17}-3<2,$则$a = 1,$$b=\sqrt {17}-3 - 1=\sqrt {17}-4$
$(2)(-3a)^3+(b + 4)^2=(-3×1)^3+(\sqrt {17}-4 + 4)^2=-27 + 17=-10$
∴$(-3a)^3+(b + 4)^2$的立方根为$\sqrt [3]{-10}=-\sqrt [3]{10}$