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.
You are close. Here are a few things that should help. First, the valuation of the different is the minimum of the terms $v_{\pi}(a_i) + (p-i)$; not the maximum.
Second, you need the extension $L/\mathbb{Q}_p$ to be a totally ramified Galois extension. The two results you mention are not necessarily true if the extension is merely totally ramified. For example, consider the extension of $\mathbb{Q}_p$ defined by the Eisenstein polynomial $x^p+px+p$; the valuation of the different is $p$.
Third, for a totally ramified extension of $\mathbb{Q}_p$ of degree $n$ with uniformizer $\pi$ and different $D$, it's important to remember that $v_{\pi}(D)$ is bounded below by $n-1$ and above by $v_{\pi}(n)+(n-1)$; you can prove this by analyzing $v_{\pi}(f'(\pi))$ for Eisenstein polynomials $f(x)$. The upper bound is helpful in establishing the two results you mention. If $n=p$, the upper bound is $2p-1$. If $n=p^2$, the upper bound is $3p^2-1$.
Fourth, let $G$ be the Galois group of $L/\mathbb{Q}_p$ and $G_i$ the $i$-th ramification subgroup of $G$. It is helpful to remember that the integers where $G_i \neq G_{i+1}$ are congruent mod $p$ and, in these cases, $\#(G_i/G_{i+1}) = p$. See Serre's book on Local Fields for proofs, if desired.
To prove item 1, you are correct that $\#G = \#G_0 = \#G_1 = p$, since the extension is totally ramified. If it were the case that $\#G_2 = p$, then $v_{\pi}(D) = \sum_{i\geq0} (\#G_i-1) \geq 3(p-1) = 3p-3$. This is a contradiction since $p$ is odd and the upper bound for $v_{\pi}(D)$ is $2p-1$; by my third comment above. Thus $\#G_i = 1$ for $i\geq 2$, establishing the result.
Proving item 2 is similar in spirit. As before, $\#G = \#G_0 = \#G_1 = p^2$. We reach a contradiction if $\#G_2 = p^2$. To see this, notice that we will have $\#G_i\geq p$ for $3\leq i\leq p+2$ (by my fourth comment above). Therefore $v_{\pi}(D) = \sum_{i\geq0} (\#G_i-1) \geq 3(p^2-1)+p(p-1) = 4p^2-p-3$, which is larger than the upper bound of $3p^2-1$ since $p$ is odd.
It follows that $\#G_2 = p$, and thus $\#G_i = p$ for $2\leq i\leq p+1$ (again by my fourth comment above). If it were the case that $\#G_{p+2} = p$, this would then imply that $\#G_i = p$ for $2\leq i \leq 2p+1$. Therefore $v_{\pi}(D) = \sum_{i\geq0} (\#G_i-1) \geq 2(p^2-1)+2p(p-1) = 4p^2-2p-2$, which is again larger than the upper bound of $3p^2-1$ since $p$ is odd. Thus $\#G_i = 1$ for $i\geq p+2$, and this establishes the result.
Best Answer
If $n$ is prime, then any extension will be either unramified or totally ramified, because $n=ef$. Suppose that $n$ is not prime and $n=ml$ with $m,l>1$. Take a degree $m$ totally ramified extension, say $\Bbb Q_p(p^{1/m})$ and a degree $l$ unramified extension $\Bbb Q_p(\zeta_{w})$ where $w=p^l-1$. Then we can take the composite field $\Bbb Q_p(p^{1/m},\zeta_{w})$ this has degree $ml=n$. Indeed $x^m-p$ is is still Eisenstein over the field $\Bbb Q_p(\zeta_w)$ because $p$ is a prime element in the ring of integers (as $\Bbb Q_p(\zeta_w)/\Bbb Q_p$ is unramified). Because the ramification index and inertia degree are multiplicative, this extension has ramification index $m$ and inertia degree $l$, so it's ramified, but not totally ramified.