Let $p$ be a prime number, $\overline{\mathbb{Q}_p}$ a fixed algebraic closure of $\mathbb{Q}_p$.
Let $\alpha$ be a primitive $n$-th root of unity in $\overline{\mathbb{Q}_p}$ and $d$ its degree over $\mathbb{Q}_p$. I know that $d=p-1$ when $n=p$ (I guess also $d=p^{k-1}(p-1)$ when $n=p^k$, but I am not sure) and $d=f$ when $n=p^f-1$.
How do one deduce $d$ in general? Is there a thing like a "$p$-adic Euler's totient function"? What are its known properties? Is there a relation with the classical Euler's totient function?
All you need to do is to put the pieces you listed together. Let $n=p^kn_1$, where $\gcd(n_1,p)=1$, and $\zeta$ be a primitive root of unity of order $n$. Then $$ \Bbb{Q}_p(\zeta)=\Bbb{Q}_p(\zeta^{n_1},\zeta^{p^k}). $$ Here $\zeta^{n_1}$ is a root of unity of order $p^k$ so the intermediate extension $\Bbb{Q}_p(\zeta^{n_1})/\Bbb{Q}_p$ is a totally ramified extension of degree $p^{k-1}(p-1)$. On the other hand $\zeta^{p^k}$ is a root of unity of order $n_1$, so the intermediate extension $\Bbb{Q}_p(\zeta^{p^k})/\Bbb{Q}_p$ is unramified, and of degree $f$, where $f$ is the smallest positive integer with the property $n_1\mid p^f-1$.
Because one extension is totally ramified and the other is unramified, the two extensions are linearly disjoint, and hence the degree of the compositum is $$ [\Bbb{Q}_p(\zeta):\Bbb{Q}_p]=p^{k-1}(p-1)f. $$
I would summarize this as follows. The roots of unity of $p$-power order behave as they do over $\Bbb{Q}$, and give rise to totally unramified extensions (ramification is a local phenomenon). The roots of unity of $p$-prime order come in lumps of factors of $p^f-1$ - all of them come together. This matches exactly the behavior of extensions of finite fields $\Bbb{F}_{p^f}/\Bbb{F}_p$, where the bigger field consists of all the roots of unity of order (a factor of) $p^f-1$. Hensel's lemma gives the bridge from the finite to the $p$-adics as it allows us to lift the modulo $p$ factorization of the cyclotomic polynomial $\Phi_{n_1}(x)$ to the $p$-adic domain.