[sp]The triple $(m,n,k) = (2,1,1)$ satisfies $m^3+n^3+k^3 = 5mnk$. Since those three lengths do not form a triangle, it follows that $p$ cannot be greater than $5$. To show that $p=5$ does have the given property, it will be sufficient to show that every triple $(m,n,k)$ of lengths that do not form a triangle must satisfy the inequality $m^3 + n^3 + k^3 \geqslant 5mnk.$
We may assume that $k$ is the largest of the three numbers. Then the condition that $m$, $n$, $k$ do not form a triangle is $m+n\leqslant k$. The inequality $m^3 + n^3 + k^3 \geqslant 5mnk$ is homogeneous of degree $3$, so it will be sufficient to scale $m$, $n$, $k$ so that $k=1$ and $m+n\leqslant1$. The inequality then becomes $5mn - m^3 - n^3 \leqslant 1.$ Thus we want to find the maximum value of $f(m,n) = 5mn - m^3 - n^3$ in the triangle $m\geqslant 0$, $n\geqslant 0$, $m+n\leqslant 1.$
The function $f(m,n)$ has no critical points inside the triangle (in fact, its only critical points are $(0,0)$ and $\bigl(\frac53,\frac53\bigr)$), so its maximum value must occur on the boundary. If $m=0$ or $n=0$ then $f(m,n)=0$. If $m+n=1$ then $$f(m,n) = 5m(1-m) - m^3 - (1-m)^3 = -8m^2 + 8m - 1 = 1 - 2(2m-1)^2,$$ which has a maximum value $1$ (when $m= \frac12$). Therefore $f(m,n)\leqslant 1$ in the triangle, as required.[/sp]