Books on algebraic number theory should have this.
For $n$ not divisible by $p$, the Galois group of $\mathbb{Q}_p(\zeta_n)$ over $\mathbb{Q}_p$ is the same as the Galois group of $\mathbb{F}_p(\zeta_n)$ over $\mathbb{F}_p$, basically because of Hensel's Lemma. But finite extensions of finite fields are cyclic, so it's easy to calculate the latter Galois group: It's cyclic of size equal to the multiplicative order of $p$ modulo $n$. The extension is also unramified.
For $n=p^k$, the Galois group of $\mathbb{Q}_p(\zeta_n)$ over $\mathbb{Q}_p$ is $(\mathbb{Z}/p^k \mathbb{Z})^{\times}$ , same as over $\mathbb{Q}$. For $k=1$, you can see directly that the cyclotomic polynomial $(x^p-1)/(x-1)$ is irreducible because it becomes an Eisenstein polynomial upon making the change of variable $x \mapsto x+1$. [Edit: This substitution doesn't always work for higher $k$; I'll have to think of another argument here.] The extension is totally ramified.
For general $n$, combine the previous 2 cases.
Somehow, the theme of symmetrization often doesn't come across very clearly in many expositions of Galois theory. Here is a basic definition:
Definition. Let $F$ be a field, and let $G$ be a finite group of automorphisms of $F$. The symmetrization function $\phi_G\colon F\to F$ associated to $G$ is defined by the formula
$$
\phi_G(x) \;=\; \sum_{g\in G} g(x).
$$
Example. Let $\mathbb{C}$ be the field of complex numbers, and let $G\leq \mathrm{Aut}(\mathbb{C})$ be the group $\{\mathrm{id},c\}$, where $\mathrm{id}$ is the identity automorphism, and $c$ is complex conjugation. Then $\phi_G\colon\mathbb{C}\to\mathbb{C}$ is defined by the formula
$$
\phi_G(z) \;=\; \mathrm{id}(z) + c(z) \;=\; z+\overline{z} \;=\; 2\,\mathrm{Re}(z).
$$
Note that the image of $\phi$ is the field of real numbers, which is precisely the fixed field of $G$. This example generalizes:
Theorem. Let $F$ be a field, let $G$ be a finite group of automorphisms of $F$, and let $\phi_G\colon F\to F$ be the associated symmetrization function. Then the image of $\phi_G$ is contained in the fixed field $F^G$. Moreover, if $F$ has characteristic zero, then $\mathrm{im}(\phi_G) = F^G$.
Of course, since $\phi_G$ isn't a homomorphism, it's not always obvious how to compute a nice set of generators for its image. However, in small examples the goal is usually just to produce a few elements of $F^G$, and then prove that they generate.
Let's apply symmetrization to the present example. You are interested in the field $\mathbb{Q}(\zeta_7)$, whose Galois group is cyclic of order $6$. There are two subgroups of the Galois group to consider:
The subgroup of order two: This is the group $\{\mathrm{id},c\}$, where $c$ is complex conjugation. You have already used your intuition to guess that $\mathbb{Q}(\zeta_7+\zeta_7^{-1})$ is the corresponding fixed field. The basic reason that this works is that $\zeta_7+\zeta_7^{-1}$ is the symmetrization of $\zeta_7$ with respect to this group.
The subgroup of order three: This is the group $\{\mathrm{id},\alpha,\alpha^2\}$, where $\alpha\colon\mathbb{Q}(\zeta_7)\to\mathbb{Q}(\zeta_7)$ is the automorphism defined by $\alpha(\zeta_7) = \zeta_7^2$. (Note that this indeed has order three, since $\alpha^3(\zeta_7) = \zeta_7^8 = \zeta_7$.) The resulting symmetrization of $\zeta_7$ is
$$
\mathrm{id}(\zeta_7) + \alpha(\zeta_7) + \alpha^2(\zeta_7) \;=\; \zeta_7 + \zeta_7^2 + \zeta_7^4.
$$
Therefore, the corresponding fixed field is presumably $\mathbb{Q}(\zeta_7 + \zeta_7^2 + \zeta_7^4)$.
All that remains is to find the minimal polynomials of $\zeta_7+\zeta_7^{-1}$ and $\zeta_7 + \zeta_7^2 + \zeta_7^4$. This is just a matter of computing powers until we find some that are linearly dependent. Using the basis $\{1,\zeta_7,\zeta_7^2,\zeta_7^3,\zeta_7^4,\zeta_7^5\}$, we have
$$
\begin{align*}
\zeta_7 + \zeta_7^{-1} \;&=\; -1 - \zeta_7^2 - \zeta_7^3 - \zeta_7^4 - \zeta_7^5 \\
(\zeta_7 + \zeta_7^{-1})^2 \;&=\; 2 + \zeta_7^2 + \zeta_7^5 \\
(\zeta_7 + \zeta_7^{-1})^3 \;&=\; -3 - 3\zeta_7^2 - 2\zeta_7^3 - 2\zeta_7^4 - 3\zeta_7^5
\end{align*}
$$
In particular, $(\zeta_7+\zeta_7^{-1})^3 + (\zeta_7+\zeta_7^{-1})^2 - 2(\zeta_7+\zeta_7^{-1}) - 1 = 0$, so the minimal polynomial for $\zeta_7+\zeta_7^{-1}$ is $x^3 + x^2 - 2x - 1$. Similarly, we find that
$$
(\zeta_7 + \zeta_7^2 + \zeta_7^4)^2 \;=\; -2 - \zeta_7 - \zeta_7^2 - \zeta_7^4
$$
so the minimal polynomial for $\zeta_7 + \zeta_7^2 + \zeta_7^4$ is $x^2+x+2$.
Best Answer
First of all, the formulation doesn't make much sense; surely you want to ask for an extension $L/\mathbf{Q}$ such that the inclusion $K \subset L$ induces a surjection
$$G = \mathrm{Gal}(L/\mathbf{Q}) \rightarrow \mathrm{Gal}(K/\mathbf{Q}) = H.$$
There is then a second ambiguity as to whether you want to fix the choice of this map (note that the last isomorphism is canonical so this makes a difference: if $K = \mathbf{Q}(\sqrt{-1},\sqrt{5})$ so $\mathrm{Gal}(K/\mathbf{Q}) = (\mathbf{Z}/2 \mathbf{Z})^2$ and if $$G = (\mathbf{Z}/2) \oplus (\mathbf{Z}/4)$$ then two of the three surjective maps will not correspond to fields $L$ but the third does. Of course the question with the map unspecified is just the question ranging over the finitely many choices of map.
The typical way to study (central) extensions is via class field theory and the Brauer Group, but since everything is in the abelian context that is not really necessary. Instead, make some reductions:
Because abelian groups are canonically direct sums of their $p$-Sylow subgroups, you can immediately reduce to the case where $H$ and $G$ have $p$-power order.
If $G = G' \oplus \Gamma$ where the map $G \rightarrow H$ factors through $G'$, you can replace $G$ by $G'$. That is because if $L'/K$ is an extension with Galois group $G'$, you can always find a disjoint abelian extension with Galois group $\Gamma$ and then take the compositum. From basic facts about $p$-groups (or even more basic facts about abelian groups), this means that we can assume that $G$ and $H$ have the same number of generators, and that $G/p=H/p$.
Explicitly, one should think about the case
$$G = \prod \mathbf{Z}/p^{a_i + b_i} \mathbf{Z}, H = \prod \mathbf{Z}/p^{a_i} \mathbf{Z}.$$
In the version of the problem where the map is specified, we can even reduce to the case where $G$ and $H$ are cyclic. (This requires a small lemma: if $L_i/K_i$ are the extensions, then one wants to make sure that the $L_i$ are all disjoint, since otherwise they only give rise to a group $G'$ surjecting onto $H$ with $G' \subset G$. But any non-trivial intersection of the $L_i$ would contain a field of degree $p$, but the degree $p$ subfields of their compositum correspond to quotients of $G'/p = H/p$ (since $G/p = H/p$), and thus one is OK since the $K_i$ are disjoint.
This paragraph now contains the key claim: If an $L$ exists (with $G/p = H/p$) then an $L$ exists which has the following property:
$L/\mathbf{Q}$ is only ramified at the (finite) primes where $K/\mathbf{Q}$ is ramified.
We now prove this. It suffices to consider the cyclic case $L = L_i$ and $K = K_i$, since if all the $L_i$ are unramified at $q$ then so is their compositum. So we have a surjection:
$$\mathbf{Z}/p^{a+b} \mathbf{Z} = \mathrm{Gal}(L/\mathbf{Q}) \rightarrow \mathrm{Gal}(K/\mathbf{Q}) = \mathbf{Z}/p^b \mathbf{Z}.$$
Suppose that the image of inertia at the prime $q$ has order $p^r$. That means that, for a prime $v$ above $q$, the extension $K_v/\mathbf{Q}_q$ is ramified of degree $p^r$. But local class field theory implies that every unramified extension of $\mathbf{Q}_q$ is contained in extension of $\mathbf{Q}_q$ containing all $q$-power roots of unity. Moreover, this extension can be "globalized," that is, there exists an extension $E/\mathbf{Q}$ of the same degree $p^r$ which is contained in the $q^{\infty}$ roots of unity and such that $\mathrm{Gal}(E/\mathbf{Q})$ is cyclic of order $p^r$. So now consider the compositum of $L$ and $E$. Note that $L$ and $E$ are disjoint, because any intersection would contain a degree $p$ extension, but (since $L$ is cyclic) it has a unique such extension which is contained in $K$ and $K$ is unramified at $q$. So now
$$\mathrm{Gal}(L.E/\mathbf{Q}) = \Gamma = \mathbf{Z}/p^{a+b} \mathbf{Z} \oplus \mathbf{Z}/p^{r} \mathbf{Z},$$
and the inertia group $I$ at $q$ is still cyclic of order $p^r$, because $L$ and $E$ locally give the same type of ramified extension. Moreover, the map to $\mathrm{Gal}(K/\mathbf{Q})$ is projection onto the second factor. But this map factors through the quotient by $I$ because $K$ is unramified. And, by construction, the quotient by $I$ is cyclic of order $p^{a+b}$ (generated by $[(1,0)]$). Thus we can replace $L$ by an extension $L'$ which is ramified at exactly the same primes except for $q$. Repeat with all primes where $L$ is ramified by $K$ is not. This is the crossing with a cyclic extension which gets used in many places, in particular in the proof of the Kronecker-Weber theorem.
Now using the Kronecker-Weber theorem, if $K$ is ramified only at primes dividing $N$, then $K \subset \mathbf{Q}(\zeta_N)$ if $(N,p) = 1$ or $\mathbf{Q}(\zeta_M,\zeta_{p^{\infty}})$ if $N=Mp$. Hence $H$ is canonically a quotient of $$\mathbf{Z}^{\times}_p \times \prod_{q|M} (\mathbf{Z}/q \mathbf{Z})^{\times}$$ if $N = pM$ or $$\prod_{q|M} (\mathbf{Z}/q \mathbf{Z})^{\times}$$ otherwise.
So now for $G$ (with the same number of generators) to arise, it is sufficient and necessary that $G$ is also a quotient of this group. (Clearly you can specify the map to $H$ or not if you like.) Of course, you need to understand cyclotomic fields enough to understand what the specific map from the group above to $H$ is (slightly annoying in the case when $p = q = 2$.
Examples 1: $H$ cyclic of order $p$, and $G$ cyclic of order $p^{2}$.
Then such an $L$ exists if and only if all of the following are satisfied.
$$d = -1, -5, 2, 10, -2, -10, 5.$$
The last one is unramified and so doesn't occur in this context, and otherwise the condition is that the localization of $K$ at the prime $2$ does not contain (equal) the cases $d = -1, -5, -2, -10$, i.e. it is $d = 2$ or $d = 10$.
Example 2: If $H$ is cyclic of order $p^a$ and $G$ is cyclic of order $p^{a+b}$ with $b > 0$ (pretty much the general case) then you win if and only if: