I know that there is classification of local fields, but here is a closely related question: Can the additive group of $\mathbb{Q}$ be a proper dense subgroup of a locally compact abelian group, whose topology is complete, other than the p adic numbers or the reals? I think of this question more as a collection, and I guess I will have to try out various examples here.
1.Example by MattE: Consider $\alpha =(\alpha_1, \dots, \alpha_n) \in \mathbb{R}^n$ linearly independent over $\mathbb{Q}$, then the map $q \mapsto q \alpha:= ( q \alpha_1, \dots, q \alpha_n)$ becomes dense in the $n$ torus, i.e. $\mathbb{R}^n / \mathbb{Z}^n$, actually even more it becomes equidistributed in the following sense $$\frac{1}{N}\sum\limits_{n \leq N} f(n \alpha) \rightarrow \int\limits_{\mathbb{R^N} / \mathbb{Z}^n} f( x) \mathrm{d} x.$$
It can surely be embedded densely in many such groups. E.g. it can be embedded into $(S^1)^n$ for any $n$. (Here $S^1$ is the circle group.)
To see this, choose an element $\alpha$ in $(S^1)^n$ whose powers are dense in $(S^1)^n$.
Now for inductively, for each integer $m$, choose $\alpha_m$ such that $\alpha_m^m = \alpha$, in a compatible way (i.e. so that if $m' = d m,$ then $\alpha_{m'}^d = \alpha_m$). Then the $\alpha_m$ together generate a copy of $\mathbb Q$ inside $(S^1)^n$, which will be dense.
(A little more succintly, I am using the fact that $(S^1)^n$ is divisible, hence injective, to extend the embedding $\mathbb Z \hookrightarrow (S^1)^n$ to an embedding $\mathbb Q \hookrightarrow (S^1)^n$.)
Another way to think about this example, when $n = 2$ say, is that we take a line with irrational slope in $(S^1)^2$; this gives a dense copy of $\mathbb R$, which contains inside it a dense copy of $\mathbb Q$.