Let me make a stab at this, even though I’m hardly an expert.
We have $\Bbb Q_p\subset\overline{\Bbb Q_p}\subset\widehat{\overline{\Bbb Q_p}}=\Bbb C_p$. The second field is an algebraic closure of $\Bbb Q_p$, and the third is the completion of that, known to be algebraically closed (by a theorem).
The second field, $\overline{\Bbb Q_p}$ has a canonical absolute value $|\star|_p$ extending the standard absolute value on $\Bbb Q_p$. The only role of Axiom of Choice is in the construction of the algebraic closure, not in defining the absolute value on it.
From your question, I divine that you are not familiar with the definition of the absolute value on $\overline{\Bbb Q_p}$. To talk about it, I will (partly for typographical reasons) use the corresponding additive valuation $v_p(z)$, defined to be $-\log_p\bigl(|z|_p\bigr)$, so that $v_p(p)=1$, $v(zw)=v(z)+v(w)$, and $v(z+w)\ge\min\bigl(v(z),v(w)\bigr)$. The valuation of $0$ is set at $\infty$. Now, for every finite extension $K\supset\Bbb Q_p$, there is a unique extension of $v_p$ from $\Bbb Q_p$ to $K$, namely
$$
v_K(z)=\frac1{[K:\Bbb Q_p]}v_{\Bbb Q_p}\bigl(\mathbf N^K_{\Bbb Q_p}(z)\bigr)\,,
$$
where the $\mathbf N^K_{\Bbb Q_p}$ is the field-theoretic norm from $K$ down to $\Bbb Q_p$. To show that this is a valuation, you need the properties of the norm, completeness, and Hensel’s Lemma.
Now $\overline{\Bbb Q_p}$ is the union (more precisely the direct limit with respect to inclusions) of the finite extensions $K$ of $\Bbb Q_p$. In particular, to find $v(z)$ for $z\in\overline{\Bbb Q_p}$, you find a finite extension $K$ of that contains $z$, and calculate $v$ of that, using your handy-dandy formula.
So that’s the absolute value on $\overline{\Bbb Q_p}$, which is not complete under this absolute value. For that, you have to pass to the completion, $\Bbb C_p$, whose elements are, in the usual way, represented by Cauchy sequences of elements of the algebraic closure.
Now comes the place where people can get confused; I know I was confused about this for years, back when I was in school. The function $v$ takes its values, most properly, in $\Bbb Q\cup\{\infty\}$, the infinite value corresponding to the zero-value of the absolute-value function. Here’s why: From the nature of the valuation as non-Archimedean, that is $v(z+w)\ge\min\bigl(v(z),v(w)\bigr)$, a Cauchy sequence $\{z_n\}$ whose limit is not zero has the sequence $\{v(z_n)\}$ becoming eventually constant. This constant, of course, will be in $\Bbb Q$.
In other words, to get the absolute value of an element $\xi\in\Bbb C_p$, you do not need to do any limiting process beyond knowing something about a sequence whose limit is $\xi$.
Let me give you an example, first of a series whose limit is not a $\Bbb C_p$-element, then one whose limit is. The series $\sum_1^\infty p^{1/n}$ is not $p$-adically Cauchy, because the individual terms don’t get small: $v(p^{1/n})=1/n$, not approaching $\infty$, (i.e. the absolute values don’t go to zero). But the series $\sum_1^\infty p^{n\,+\,1/n}$ is Cauchy, with terms getting small (valuations going to infinity), and so represents a good element of $\Bbb C_p$. And its absolute value is that of the very first term, $p^2$, whose valuation is $2$ and whose absolute value is $\frac1{p^2}$.
For the completion of an algebraic closure of $k((t))$, for a field $k$ of characteristic $p>0$, the story is precisely the same. It is surpassingly hard to understand just the algebraic closure, at least it is for me, but the same formulas work just fine, and you’re good to go.
Yes, $\vert x\vert$ means the outcome is always non-negative. This requires clarification because $x$ itself may be negative. Here are a few examples:
$$\sqrt{\color{blue}{+4}^2} = \sqrt{16} = \color{blue}{+4}$$
$$\sqrt{\color{blue}{+9}^2} = \sqrt{81} = \color{blue}{+9}$$
$$\sqrt{(\color{blue}{-4}^2)} = \sqrt{16} = \color{blue}{+4} = -(\color{blue}{-4})$$
$$\sqrt{(\color{blue}{-9}^2)} = \sqrt{81} = \color{blue}{+9} = -(\color{blue}{-9})$$
So, when $x > 0$, then $\sqrt{x^2}$ will return $x > 0$, so it equals $x$. However, when $x < 0$, then $\sqrt{x^2}$ will return $x > 0$ as well, so it equals $-x$. Hence,
$$\sqrt{x^2} = \vert x\vert = \begin{cases} x; \quad x \geq 0 \\ -x; \quad x < 0 \end{cases}$$
This is all just a way of saying $\sqrt{x^2}$ will always return a non-negative value.
Best Answer
In $\mathbb{R}$, the absolute value function may seem too simple to be useful. But the 'idea' of an absolute value is generalizable and quite important, because it captures the concept of distance between two points. For example, in $\mathbb{R}$, $\lvert x-y \rvert$ tells us how far $x$ is from $y$. It measures distance. Now move up to $\mathbb{R}^3$. We can grasp the idea of distance between points $x$ and $y$ in $\mathbb{R}^3$, but how do we denote it? We can write $\lvert x-y \rvert$. Now, this again gives the distance between $x$ and $y$, but it is not the same simple function as it was in $\mathbb{R}$; however, it captures the same idea.
Having said that, even in $\mathbb{R}$, I would argue that the absolute value function simplifies notation a lot. For example, when we talk about a sequence $a_n$ converging to a limit $a$, we usually say something like $\forall \epsilon >0, \exists N$ such that if $n>N$, $\lvert a_n - a \rvert < \epsilon$. If we didn't use the absolute value function, we would have written $a_n - a < \epsilon$ if $a_n > a$ and $a - a_n < \epsilon$, otherwise. Obviously, this is more cumbersome.