The short answer is (as far as I am aware) no, but there is a lot that is known. Jannsen and Wingberg have given an explicit presentation for $Gal(\overline{K}/K)$ in the case that the residue characteristic is not $2$ (published in Inventiones Math in 1982/1983), and Volker (1984, Crelle) handles the case when $K$ has residue characteristic $2$ and $\sqrt{-1} \in K$. This does not, however, make it trivial to determine which finite groups are quotients of $Gal(\overline{K}/K)$. Some more information can be obtained from Section VII.5 of "Cohomology of Number Fields" by Neukirch, Schmidt and Wingberg. Here's a paraphrase.

If $K$ is a local nonarchimedean field with residue field of characteristic $p$ (and order $q$), let $G = Gal(\overline{K}/K)$, $T$ be the inertia group, and $V$ be the ramification group. Then $G/T \cong \hat{\mathbb{Z}}$, $T/V \cong \prod_{\ell \ne p} \mathbb{Z}_{\ell}$, and $V$ is a free pro-$p$ group of countably infinite rank. Iwasawa showed that $G/V$ is a profinite group with two generators $\sigma$ and $\tau$ so that $\sigma \tau \sigma^{-1} = \tau^{q}$. Also, the maximal pro-$\ell$ quotient of $G$ is known for all $\ell$. For example, if $\mu_{\ell} \not\subseteq K$ and $\ell \ne p$, the maximal pro-$\ell$ quotient is $\mathbb{Z}_{\ell}$ (i.e. for each positive integer $k$, there is a unique Galois extension $L/K$ of degree $\ell^{k}$, namely the unramified one).

1more comment