[Edit to answer the modified question:]
For $p=1$, $\epsilon_{n,p}(m)=(n-1)m$ and there is no upper bound on $\epsilon$, so I will assume $p>1$.
The derivative of this new version of $\epsilon$ with respect to $m$ is
\epsilon'_{n,p}(m) = n^{1/p}m^{-(p+1)}\left(p - 1 + m^{-p}\right)^{-(p+1)/p} - 1\;.
Setting this to zero yields
$n^{1/p}m^{-(p+1)}\left(p - 1 + m^{-p}\right)^{-(p+1)/p}=1\;,$
$n^{-1/(p+1)}m^{p}\left(p - 1 + m^{-p}\right)=1\;,\tag{1}$
$(p - 1)m^p + 1=n^{1/(p+1)}\;,$
$m=\left(\frac{n^{1/(p+1)}-1}{p-1}\right)^{1/p}\;.\tag{2}$
Then using (1) to substitute $p - 1 + m^{-p}$ and (2) to substitute $m$ yields
$ \begin{eqnarray} \epsilon_{n,p}(m) &=& \left(n^{1/p}n^{-1/(p(p+1))}-1\right)m \\ &=& \left(n^{1/(p+1)}-1\right)\left(\frac{n^{1/(p+1)}-1}{p-1}\right)^{1/p} \\ &=& \left(\frac{\left(n^{1/(p+1)}-1\right)^{p+1}}{p-1}\right)^{1/p}\;. \end{eqnarray} $
For $n=3$, $p=3$, this is about $0.171$, in agreement with your plot. To show that this is always a global maximum, note that $\epsilon_{n,p}(m)$ goes to $0$ for $m\to0$ and to $-\infty$ for $m\to\infty$, so if the derivative vanishes at a single point, it must be a global maximum.