For $n=3$ we need to find $$\max_{a^2+b^2+c^2=1}(a^2b+b^2c+c^2a).$$
Indeed, let $\{|a|,|b|,|c|\}=\{x,y,z\}$, where $x\geq y\geq z\geq0$.

Thus, by Rearrangement and AM-GM we obtain:
$$\sum_{cyc}a^2b\leq|a|\cdot(|a||b|)+|b|\cdot(|b||c|)+|c|\cdot(|c||a|)\leq$$
$$\leq x\cdot xy+y\cdot xz+z\cdot yz=y(x^2+xz+z^2)\leq y\left(x^2+\frac{x^2+z^2}{2}+z^2\right)=$$
$$=\frac{3}{2}y(1-y^2)=\frac{3}{2\sqrt2}\sqrt{2y^2(1-y^2)^2}\leq\frac{3}{2\sqrt2}\sqrt{\left(\frac{2y^2+2-2y^2}{3}\right)^3}=\frac{1}{\sqrt3}.$$
The equality occurs for $a=b=c=\frac{1}{\sqrt3}$, which says that we got a maximal value.