The classification of 1-dimensional representations, i.e. characters $G_K \to E^\times$, is a bit more complicated than you imply in your question. Any such character lands in $O_E^\times$ by compactness; and since $O_E^\times$ is profinite, and class field theory identifies the abelianisation of $G_K$ with the profinite completion of $K^\times$, we conclude that there is a canonical bijection between continuous characters $G_K \to O_E^\times$ and continuous characters $K^\times \to O_E^\times$.

In terms of this bijection, we can read off which characters are crystalline, semistable, de Rham or Hodge--Tate using the restriction of the character to $O_K^\times$: the details are given in Appendix B of B. Conrad, "Lifting global representations with local properties"; see also this question and this question.

As for $n = 2$, the best answer I can give is that for $K = \mathbf{Q}_p$, there is a bijection between 2-dimensional representations of $G_{\mathbf{Q}_p}$ and certain $p$-adic Banach space representations of $GL_2(\mathbf{Q}_p)$: this is the $p$-adic local Langlands correspondence of Colmez. You can then classify which 2-dimensional reps are de Rham / crystalline in terms of their associated $GL_2$ representation. However, this is a bijection between one kind of formidably complicated object and another kind of equally complicated object; there's no hope of getting a down-to-earth parametrisation of all the representations involved. Moreover, extending this correspondence to 2-dimensional reps of $G_K$, for arbitrary $K$, or to 3-dimensional reps of $G_{\mathbf{Q}_p}$, is an open problem despite a decade or more of very intensive effort.

(Another viewpoint: if you just want to classify crystalline / semistable / de Rham reps, but you don't necessarily ask for a classification of *all* reps and how the crys/ss/dR ones sit inside that, then you can do this in some cases by classifying all filtered $\varphi$-modules, $(\varphi, N)$-modules, etc satisfying the relevant conditions. See e.g. this paper of Dousmanis. But this tells you nothing about what the non-de Rham guys look like.)