Degree of extension concerning p-adic numbers. We define
$$\mathbb{Z}[[X]]_{conti}:=\{t:\mathbb{R}_{\geq 0}\rightarrow \mathbb{Z}\ (t\in \mathbb{Z}^{\mathbb{R}_{\geq 0}})|\forall M\geq 0\ \{r\in\mathbb{R}_{\geq0}|r\leq M\land t(r)\neq0\}\ are \ finite\ sets\}.$$
Let $\mathbb{Q}_p^{ext}$ be $\mathbb{Z}((X))_{conti}/(X-p)$ for prime number $p$.
Then, what is the degree of extension of algebraic closure over $\mathbb{Q}_p^{ext}$?
 A: I guess you meant $$A=\Bbb{Z}[[X]]_{conti}/(X-p),\qquad K=Frac(A)$$
The elements of $A$ are formal series $\sum_{j\ge 0} c_j p^{e_j}$ with $c_j\in 0\ldots p-1$ and $e_j\in \Bbb{R}_{\ge 0}$ strictly increasing and $e_j\to \infty$,
with addition and multiplication defined similarly to the ones on the $p$-adic series definition of $\Bbb{Z}_p$.
$K$ is the field of formal series $\sum_{j\ge 0} c_j p^{e_j}$ with $e_j$ allowed to include finitely many negative terms, ie. $K=A[p^{-1}]$.

This is the first time I see those rings, interesting for sure.

There is a natural non-archimedian valuation on $K$, namely $$v(\sum_{j\ge 0} c_j p^{e_j})=\inf_{c_j\ne 0} e_j$$
The valuation ring is $A$ and the residue field is $\Bbb{F}_p$, whose algebraic closure is an infinite extension.
So $[\overline{K}:K]$ is infinite.
Is there a chance that $K\otimes_{\Bbb{Q}_p[p^{1/\infty}]} \overline{\Bbb{Q}}_p$ is a field and algebraically closed? If it is algebraically closed then $[\overline{K}:K]$ would be countably infinite.
