This is an exercise Iāve never done, but it should be a lot of fun. What is the general Eisenstein polynomial in this case? itāll 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 doesnāt 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. Itās a nice question, though, and I think Iām going to worry it over a little.
EDIT ā Expansion:
I told no lies above, but thatās 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, thereās 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 $X'=X-2a/3$, youāll get a new Eisenstein polynomial, but without a quadratic term. Now, if you calculate the discriminant of $X^3+2bX+2(1+2c)$, youāll 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 $k$. Thatās Kummer Theory, as Iām sure you know.
Letās call $\Bbb Z_2[\omega]=\mathfrak o$, thatās 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, thatās 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 thatās 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.
$F=K((t))$ with $K$ algebraically closed of characteristic $0$.
For a finite extension $E/F$, with corresponding complete DVR $O_F=K[[t]]\subset O_E$ and uniformizers $\pi_F=t,\pi_E$ and residue field $O_F/(\pi_F)=O_E/(\pi_E)=K$.
The completeness says that $$O_E=\sum_{n\ge 0} \pi_E^n K=\sum_{n= 0}^{[E:F]-1} \pi_E^n O_F$$
ie. $\pi_F = u \pi_E^{[E:F]}$ where $u\in O_F^\times$.
Let $a\in K^\times$ such that $au \in 1+\pi_F O_F$.
Then $(au)^{1/[E:F]}=\sum_{k\ge 0} {[E:F]\choose k} (au-1)^k \in O_F$ and $a^{1/[E:F]}\in O_F$ which means that $u^{1/[E:F]}\in O_F$ ie. $\pi_F^{1/[E:F]} \in O_F$.
But $\pi_F^{1/[E:F]} $ is also an uniformizer so that $$O_E=\sum_{n\ge 0} (\pi_F^{1/[E:F]})^n K=\sum_{n= 0}^{[E:F]-1} (\pi_F^{1/[E:F]})^n O_F$$ and
$E = F(\pi_F^{1/[E:F]})$
Whence $$\overline{F}=\bigcup_{[E:F]<\infty} E=\bigcup_{[E:F]<\infty}F(\pi_F^{1/[E:F]})=\bigcup_m K((t))(t^{1/m})$$
Best Answer
There aren't too many unramified extensions of a number field. In fact, if $K$ is a number field and $K^{unr}$ is the maximal unramified extension of $K$, then $G(K^{unr}/K)\simeq Cl(K)$, where $Cl(K)$ is the ideal class group of $K$. So, for instance, there are no unramified extensions of $\mathbb{Q}$. The inverse of this map behaves like $(\mathbb{Z}/n\mathbb{Z})^{\times}\rightarrow G(\mathbb{Q}(\zeta_n)/\mathbb{Q})$, in that we send the equivalence class of a prime $[\mathfrak{p}]$ into the Frobenius of $\mathfrak{p}$. This means that a prime will split completely in $K^{unr}$ if and only if it is principle. It is possible to compute these fields, and sometimes it's easy.
If we're okay with some ramification, say ramification at all primes dividing the ideal $\mathfrak{m}$, then we can create the Ray Class Group of $\mathfrak{m}$, denoted $Cl_{\mathfrak{m}}(K)$ which is like the class group but it avoids the primes that divide $\mathfrak{m}$. We can then use a similar idea to create all extensions that are only ramified at primes dividing $\mathfrak{m}$. Ie, there is a field $K^{unr}_{\mathfrak{m}}$ that is the maximal field unramified at every prime not dividing $\mathfrak{m}$, we call this the Ray Class Field of $\mathfrak{m}$. If a prime power divides $\mathfrak{m}$, then we can control ramification at that prime.
There's a little more we can do. Sometimes a field can split over the real/complex numbers. We can control that by adding an additional, artificial term to $\mathfrak{m}$, making it a Modulus. We can then get Ray Class Fields of this modulus as well.
Class Field Theory then says that every single (abelian) field extension is a Ray Class Field of some modulus. That is good because we can compute ray class fields and categorize them. This paper gives a few algorithms that can compute these extensions.