You get unramified extensions of $\Bbb Q_p$ by adjoining roots of unity of order prime to $p$; alternatively, by adjoining $(p^n-1)$-th roots of unity.
The finite unramified extensions of $\Bbb Q_p$ are in natural one-to-one correspondence with the finite algebraic extensions of $\Bbb F_p$. This means that there’s only one unramified extension of $\Bbb Q_p$ of each degree. Want the cubic unramified extension of $\Bbb Q_3$? Find a cubic irreducible over $\Bbb F_3$, like $X^3-X-1$, and use the same polynomial’s roots over $\Bbb Q_3$.
This is an exercise I’ve never done, but it should be a lot of fun. What is the general Eisenstein polynomial in this case? it’ll be
$$
X^3 + 2aX^2+2bX+2(1+2c)\,,
$$
where $a$, $b$, and $c$ can be any $2$-adic integers. Notice that the constant term has to be indivisible by any higher power of $2$, so of form $2$ times a unit, and the units of $\Bbb Z_2$ are exactly the things of form $1+2c$. So your parameter space is $\Bbb Z_2\times\Bbb Z_2\times\Bbb Z_2$, pleasingly compact, and a general result of Krasner says that if you jiggle the coefficients a little, the extension doesn’t change. You might be able to use all this to construct your (finitely many) fields.
Not much of an answer, I know, but it was too long for a comment. It’s a nice question, though, and I think I’m going to worry it over a little.
EDIT — Expansion:
I told no lies above, but that’s not the way to look at this problem. As I reached the solution, I realized that there are really two questions here. Consider the simplest case, which you mentioned, the Eisenstein polynomial $X^3-2$. If you think of it abstractly, there’s only the one extension of $\Bbb Q_2$ here, but if you think of the subfields of some algebraically closed containing field, there are three fields, generated by $\lambda$, $\omega\lambda$, and $\omega^2\lambda$, where $\lambda$ is a chosen cube root of $2$ and $\omega$ is a primitive cube root of unity.
As usual, if you take the displayed cubic above and make a substitution $X'=X-2a/3$, you’ll get a new Eisenstein polynomial, but without a quadratic term. Now, if you calculate the discriminant of $X^3+2bX+2(1+2c)$, you’ll get $\Delta=-32b^3-27(1+4c+4c^2)$; and since $c$ and $c^2$ have same parity, we get $\Delta\equiv-3\pmod8$, definitely not a square, indeed $\sqrt\Delta\in\Bbb Q(\omega)$, hardly a surprise, I suppose. And the splitting field of our polynomial will be a cubic extension of $k=\Bbb Q_2(\omega)$, all of which we know. We need only calculate the group $k^*/(k^*)^3$, and its cyclic subgroups (of order $3$) will tell us the cubic extensions of $k$. That’s Kummer Theory, as I’m sure you know.
Let’s call $\Bbb Z_2[\omega]=\mathfrak o$, that’s the ring of integers of $k$.
To know $k^*/(k^*)^3$ we have to look at the groups $1+2\mathfrak o\subset \mathfrak o^*\subset k^*$. Now the principal units $1+2\mathfrak o$ are uniquely $3$-divisible, so no contribution to $k^*/(k^*)^3$; the next layer, $\mathfrak o^*/(1+2\mathfrak o)$ is cyclic of order $3$, generated by $\omega$, and $k^*/\mathfrak o^*$ is infinite cyclic, that’s the value group. So $k^*/(k^*)^3$ is of dimension two as an $\Bbb F_3$-vector space, and has only four one-dimensional subspaces. One is spanned by $\omega$, and its cube roots generate an unramified extension, so is not of interest to us. The other three are spanned by $2$, $2\omega$, and $2\omega^2$. ( ! )
And that’s it. Contrary to my expectation and perhaps yours, the only cubic ramified extensions of $\Bbb Q_2$ within an algebraic closure are the three I mentioned in the first paragraph of this Edit.
Best Answer
The unramified extensions of $\mathbb{Q}_p$ are very simple. There is exactly one unramified extension of $\mathbb{Q}_p$ of degree $f$ for every prime $p$ - it is the one obtained by adjoining the $p^f − 1$ roots of unity. This unramified extension is Galois and has Galois group isomorphic to $\mathbb{Z}/f\mathbb{Z}$ over $\mathbb{Q}_p$.
Among the seven quadratic extensions of $\mathbb{Q}_2$ already six are (totally) ramified, namely for $d=-1,\pm 2, -5,\pm 10$. Hence the unrafimied must be the one with $d=5$. Alternatively, let $\zeta_3=(-1+\sqrt{-3})/2$, then $\mathbb{Q}_2(\zeta_3)=\mathbb{Q}_2(\sqrt{-3})=\mathbb{Q}_2(\sqrt{5})$ is unramified by the above statement.
See also the discussion here.