Is the algebraic closure of a $p$-adic field complete

The algebraic closure has countably infinite dimension over $\mathbb Q_p$, and therefore (by the Baire category theorem) is not metrically complete. (Except the case $\mathbb Q_\infty = \mathbb R$, where the algebraic closure has finite dimension, and is metrically complete.)

explanation, added Oct 15, 2021

In $\mathbb Q_p$, let $x_n$ be a solution of $X^{n}=p$. Then $\{x_2,x_3,x_4,\dots\}$ is linearly independent over $\mathbb Q_p$. But we still need a proof that the algebraic closure does not have uncountable dimension.

former explanation
We see in Incompleteness of algebraic closure of p-adic number field, from $9$ years ago question that it is incorrect...

How about an example? In $\mathbb Q_2$, the partial sums of the series $$ \sum_{n=1}^\infty 2^{n+1/n} $$ belong to $\overline{\mathbb Q_2}$, but the sum of the series does not. The partial sums form a Cauchy sequence with no limit in $\overline{\mathbb Q_2}$.

Why does this sum not exist in $\overline{\mathbb Q_2}$ ?

It is not trivial, but interesting: any $x$ which is algebraic of degree $n$ over $\mathbb Q_2$ has a unique series expansion $$ x = \sum 2^{u_j} $$ where $u_j \to \infty$ (unless it is a finite sum) and all $u_j$ are rationals with denominator that divides $n!$. (Maybe divides $n$ in fact?) But the series expansion in this example has arbitrarily large denominators.

Small addition: One might try to repeat this process of algebraic closure / analytic completion. Kürschak proved that this process terminates after one more step, i.e. the completion of $\overline{\mathbb{Q}_p}$ is algebraically closed.

More generally, if you start with a non-archimedean valued field $K$ and denote $\overline{K}$ the algebraic closure and $\widetilde{K}$ the analytic completion, then $\widetilde{\overline{\widetilde{K}}}$ is algebraically closed.