As you say, the wild ramification group

$$\mathfrak{w}_p = \operatorname{Aut}( \overline{\mathbb{Q}_p} / \mathbb{Q}_p^{\operatorname{tame}})$$

is a pro-$p$-group. It is also topologically finitely generated, i.e., a quotient of the pro-$p$-completion of a free group of finite rank. Thus in theory we can describe this group -- i.e., $\operatorname{Aut}(\overline{\mathbb{Q}_p}/\mathbb{Q}_p)$ -- completely by giving relations.

**Added**: The above assertion that $\mathfrak{w}_p$ is topologically finitely generated is **false**, as has been pointed out to me by Chandan Singh Dalawat and Daniel Litt. The absolute Galois group of $\mathbb{Q}_p$ is finitely generated. I believe what I had in mind is that quotients of topologically finitely generated topological groups are topologically finitely generated. Which is true -- but the absolute Galois group of the maximal totally tamely ramified extension of $\mathbb{Q}_p$ is a *subgroup*, and subgroups of noncommutative topological groups do not automatically inherit finite generation. In fact it is a theorem of Iwasawa (Transactions of the AMS, 1955) that $\mathfrak{w}_p$ is a free pro-$p$ group of countably infinite rank. *Mea culpa*.

For $p > 2$, the determination of the relations for

$$\mathfrak{g}_p = \operatorname{Aut}(\overline{\mathbb{Q}_p}/\mathbb{Q}_p)$$

has been done! The set of relations is finite for each $p$ and is (for $p > 2$) explicitly known, due to work of Jannsen and Wingberg. This wikipedia article gives references.

However, to the best of my knowledge (and my knowledge here is not the best), this kind of generators and relations description has not been especially useful. (To be sure, it is a nice result. It happens to be the case that not all nice results are especially useful...) For instance, the Local Langlands Conjecture is a deep result about finite-dimensional representations of the absolute Galois group of $\mathbb{Q}_p$. So far as I know (insert further provisos...), knowing a presentation for the Galois group is utterly unhelpful in establishing results like this.

Probably M. Emerton will come along shortly and tell you more. I hope so...