jmc

By the Remainder Theorem, $p(3)=11,$ so $$A\cdot 3^5+B\cdot 3^3+C\cdot 3+4=11.$$Then $A\cdot 3^5+B\cdot 3^3+C\cdot 3=7.$

Again by the Remainder Theorem, when $p(x)$ is divided by $x+3,$ the remainder is $$\begin{array}{rl}p(-3)& =A\cdot (-3{)}^5+B\cdot (-3{)}^3+C\cdot (-3)+4\\ & =-A\cdot 3^5-B\cdot 3^3-C\cdot 3+4\\ & =-7+4\\ & =\boxed{-3}.\end{array}$$