For any field $K$ of characteristic $p$ and any non-trivial valuation $v:K\to G$ (to an ordered abelian group) then $O_v = \{ a \in K, v(a)\ge 0\}$ is a valued ring and $R=O_v^{1/p^\infty} = \{ a \in \overline{K}, \exists m, a^{p^m} \in O_v\}$ is a perfect valued ring.
Example : $O_v = \Bbb{F}_p[[x]], R= \Bbb{F}_p[[x]][x^{1/p^\infty}]$.
Any perfect valued ring is of this form.
If the maximal ideal of a valued ring is finitely generated then the value group is a finitely generated $\Bbb{Z}$-module so $a \mapsto a^p$ can't be surjective.
For the Krull dimension, the proper prime ideals of a valuation ring correspond to subsets $S$ of the value group satisfying $s > 0, nt > s \implies t \in S$. Thus the Krull dimension depends on the "depth" of the value group.
For example the non-archimedian valuation $v(x^n y^m) = n+m\epsilon\in \Bbb{Z+\epsilon Z}$ on $k[x,y]$, the order is $n +m\epsilon \ge 0$ iff $n> 0$ or $n=0,m\ge 0$, extend $v$ to $Frac(k[x,y])$ then the valuation ring $O_v = \{ a \in k(x,y), v(a) \ge 0\}= x k(y)[x]_{(x)}+k[y]_{(y)} $ has 3 prime ideals $\{0\}, (x)$ and $(x,y)$.