The finite extensions of $\Bbb Q_p$ are also local fields, so you can use the usual characterization of local field extensions being unramified.
Recall: a finite extension of local fields $L/K$ is unramified iff $[L:K]=[\lambda:\kappa]$ where $\lambda,\kappa$ are the residue fields for $L,K$ respectively. This comes from the fundamental relationship that
$$e(L|K)f(L|K)=[L:K]$$
and the fact that $f(L:K)=[\lambda:\kappa]$. But then since your $K$ is a finite extension of $\Bbb Q_p$ you know the residue field is a finite extension of $\Bbb F_p$, so that unramified extensions are classified by intermediate fields $\ell$ such that
$$\kappa\subseteq \ell\subseteq \overline{\kappa}=\overline{\Bbb F_p}$$
the last equality since the algebraic closure of $\kappa$, being itself a finite extension of $\Bbb F_p$ is the same as the algebraic closure of $\Bbb F_p$. This completely classifies unramified extensions of such $K$. You even have a nice description: if $\ell\cong \Bbb F_{p^n}$ then the corresponding extension of $K$ is just $K(\mu_{p^n-1})$ with $\mu_k$ as usual the set of $k^{th}$ roots of $1$.
From this we may further deduce a nice corollary that the maximal, unramified extension of $K$ is simply $K(\mathbf{\mu})$ with
$$\mathbf{\mu}=\bigcup_{(n,p)=1,\; \mu_n\not\subseteq K} \mu_n$$
This correspondence is--as usual--due to Hensel's lemma and the characterization of finite fields as collections of roots of unity (plus $0$). Naturally the $(n,p)=1$ condition comes from the fact that all $p^{n}$ roots of unity are already in $\kappa$ since $1$ is the only one such and the Frobenius is an automorphism.
From this you can also see that since there is no $p^n$ roots of $1$ un an unramified extension, and since $\mu_{mn}=\mu_m\times\mu_n$ when $(n,m)=1$, that any extension of $K$ by $p^n$ roots of $1$ must be totally ramified, unless $K$ already contains those roots of $1$. Wedge this against the fact that you can always write any extension as an unramified, followed by a totally ramified, and you have your more general approach.
The useful fact here is that a Dedekind domain with only finitely many primes is a PID. See, for instance, this question. The proof is so simple that I’m abashed not to have seen it.
Anyhow, your ring has only the ideals above $(p)$.
Best Answer
Suppose $\zeta$ is a primitive $n$th root of unity. The difference between $\mathbb{Q_p}[\zeta]/\mathbb{Q}_p$ and $\mathbb{Q}[\zeta]/\mathbb{Q}$ is that $\mathbb{Q}_p$ in general may contain some extra relations between the $n$th roots of unity, in which case adjoining $\zeta$ will not enlarge it as much as it does $\mathbb{Q}$.
For a general field $K$, an automorphism $\phi$ of $K[\zeta]/K$ is determined completely by where it sends $\zeta$. Since $\zeta$ is a root of unity, $\phi(\zeta)$ has to be one as well, so $\phi(\zeta)=\zeta^k$ for some integer $k$. Which integers $k$ are allowed depends on the minimal polynomial of $\zeta$, which depends on $K$.
Over $\mathbb{Q}$ the minimal polynomial of $\zeta$ is the $n$th cyclotomic polynomial $$ \Phi_n(X) = \prod_{ \substack{1 \leq k \leq n \\ gcd(k,n)=1}} (X - \zeta^k).$$
In other words the elements of ${\rm Gal}(\mathbb{Q}[\zeta]/\mathbb{Q})$ are $\phi_k(\zeta) = \zeta^k$ where $k$ belongs to $(\mathbb{Z} /n \mathbb{Z})^\times$. Then $k\mapsto \phi_k$ is an isomorphism of $(\mathbb{Z}/n\mathbb{Z})^\times$ onto ${\rm Gal}(\mathbb{Q}[\zeta]/\mathbb{Q})$ and so the latter has order $\varphi(n)$. For instance when $n=p^r$, we have $[{\rm Gal}(\mathbb{Q}[\zeta]/\mathbb{Q})]=\varphi(p^r)=p^{r-1}(p-1)$.
Now over any other field $K$ of characteristic zero, the minimal polynomial of $\zeta$ has to divide $\Phi_n(X)$. If $n=p^r$ and $K=\mathbb{Q}_p$, the same argument used to prove the irreducibility of $\Phi_{p^r}$ over $\mathbb{Q}$, in other words Eisenstein's criterion, works over $\mathbb{Q}_p$. Here it is for $r=1$.
We have $$ \Phi_{p}(X) = \frac{X^p - 1}{X-1}.$$ Setting $X = Y+1$, we can write $$ \Phi_p(X)=\Phi_p(Y+1) = \frac{(Y+1)^p - 1}{Y} = Y^{p-1} + {p \choose p-1} Y^{p-2} + \cdots + {p \choose 2}Y + p.$$
All the coefficients after the leading one are divisible by $p$. If this polynomial were reducible, so that $\Phi_p(Y+1)=Q(Y)Q'(Y)$, then modulo $p$ we would get $\overline{Q}(Y)\overline{Q}'(Y) = Y^{p-1}$ over $\mathbb{F}_p$. That would mean that $Q(Y) = Y^r$ and $Q'(Y)=Y^s$ with $s+r = p-1$. It follows that $Q(Y)$ and $Q'(Y)$ must both have constant terms divisible by $p$. But then $Q(Y)Q'(Y)$ would have a constant term divisible by $p^2$, which $\Phi_p(Y+1)$ does not. Therefore $\Phi_p(X)$ is irreducible. This is (the proof of) Eisenstein's irreducibility criterion and in this case works the same over $\mathbb{Q}_p$ as over $\mathbb{Q}$. It shows that ${\rm Gal}(\mathbb{Q}_p[\zeta]/\mathbb{Q}_p) \cong {\rm Gal}(\mathbb{Q}[\zeta]/\mathbb{Q})$.
For $n=p^r$ with $r>1$ you can use the fact that $\Phi_{p^r}(X)=\Phi_p(X^{p^{r-1}})$ and use a similar argument.
If $n\neq p^r$, $\Phi_n(X)$ may not be irreducible over $\mathbb{Q}_p$, in which case $\mathbb{Q}_p[X]/(\Phi_n(X))$ splits into a direct product of isomorphic fields, one for each distinct prime of $\mathbb{Q}[\zeta]/\mathbb{Q}$ lying over $p$. Therefore the irreducibility of $\Phi_n(X)$ over $\mathbb{Q}_p$ is tied to the splitting behaviour of the prime $p$ in $\mathbb{Q}[\zeta]$.
When $p\nmid n$, $p$ is unramified in $\mathbb{Q}_p[\zeta]/\mathbb{Q}_p$ and so ${\rm Gal}(\mathbb{Q}_p[\zeta]/\mathbb{Q}_p)$ is isomorphic to ${\rm Gal}(\mathbb{F}_p[\zeta]/\mathbb{F}_p)$, which is cyclic of order $n/gcd(n,p-1)$.