In response to Minhyong's request, I am reposting my comments above as an answer:
As James Newton commented, if $L/K$ is unramified, then an irreducible $n$-dimensional representation (over $\mathbb C$) of $Gal(L/K)$ will correspond, in the Langlands paradise, to a cuspidal automorphic representation of $GL_n(\mathbb A_K)$. The cuspidal automorphic representations that arise in this way are sometime (especially in the older literature) called "Galois type".
Thus one can (more or less --- there is the issue of irreducible vs. all reps. which I won't think about here) encode unramified extensions of $K$ whose Galois groups admit $n$-dimensional representations in terms of Galois type cuspidal automorphic representations $\pi$ of $GL_n(\mathbb A_K)$ that are unramified at every finite prime.
Now the question arises: how many such $\pi$ are there, and can one compute them?
Being of Galois type is (conjecturally, but we are in paradise!) purely a condition on $\pi_v$ for primes $v$ of $K$ lying over $\infty$, and in fact there are a finite number of prescribed representations of $GL_n(K_v)$ ($= GL_n(\mathbb R)$ or $GL_n(\mathbb C)$) which are allowed. (E.g. for $GL_n(K)$, the possibilities are limit of discrete series, corresponding to holomorphic weight one forms, or principal series with $\lambda = 1/4$, corresponding to Maass forms with eigenvalue of Laplacian equal to $1/4$.) For a given $n$ and $K$, these can be enumerated.
Now since we are asking that the "weight" (i.e. the collection of $\pi_v$ for $v|\infty$) be bounded (i.e. lie in a given finite set), and we are also asking that "level" be one (i.e. that there is no ramification at any finite prime), there are only a finite set of $\pi$ corresponding to irreducible everywhere unramified $n$-dimensional complex representations of $GL_n(Gal(\bar{K}/K)$. [Aside: Minhyong asked for a sketch of a proof of this; here goes: fixing the representation at infinity means that we are fixing a bunch of elliptic operators that the automorphic forms must satisfy. Fixing the level means that we are working on some particular quotient $X/\Gamma$ (here $X$ is the symmetric space attached to the real group in question, in our particular case $GL_n(K\otimes \mathbb R)$, and $\Gamma$ is the fixed level). This need not be compact (indeed won't be in our particular case), but the cuspidal condition (indeed, even the moderate growth condition that non-cuspidal automorphic forms are required to satisfy) means that we can pretend it is, since we explicitly rule out the possibility of extreme growth at infinity. So now we looking at sections of some bundle on a compact space satsifying a bunch of elliptic equations, and such a space of sections if finite dimensional. (The holomorphic modular forms case is the most familiar: in this case the elliptic equations are the Cauchy--Riemann equations. In the Maass form case, the corresponding fact is the finiteness of the eigenspaces of the Laplacian. These are good models for the general case.)]
To actually compute them (say for a fixed choice of $K$ and $n$) would be quite difficult (as David Hansen notes in his comment). The reason is that the relevant $\pi_v$ for $v|\infty$ are never discrete series (even when $n = 2$, and in any case, note that $GL_n(\mathbb R)$ never has discrete series if $n > 2$, and $GL_n(\mathbb C)$ never has discrete series when $n > 1$), and so standard applications of the trace formula to counting automorphic forms won't work.
Nevertheless, it seems that one might still be able to use the trace formula to analyze the situation, at least in principle. For example, Selberg used his original formulation of the trace formula for $SL_2(\mathbb R)/SL_2(\mathbb Z)$ to compute cuspidal Maass forms of level 1,
and showed that the smallest eigenvalue $\lambda$ that occurs has $\lambda$ much greater than 1/4 (maybe closer to 90?).
And we all know that it is not hard to show that there are no holomorphic weight one forms
of level one. So one can automorphically prove (modulo standard conjectures in the Maass form case) that there are no everywhere unramified two-dimensional complex representations of $Gal(\bar{\mathbb Q}/\mathbb Q)$. (This is of course an incredible battle, even in paradise, for a tiny portion of the information that Minkowski gives us, but is meant just to illustrate that this approach is not a priori ridiculous.)
What I don't see at all from this point of view is how to study all $n$ simultaneously. For example, one could imagine implementing this program and finding, for some $K$ and some $n$, maybe $n = 10^6$, that there are no unramified extensions $L/K$ with $L$ admitting an irrep. of dimension $\leq 10^6$. This doesn't rule out the possibility that there is a beautiful, everywhere unramified extension $L/K$ whose Galois group's lowest degree irrep. happens to be of enormous dimension.
The Langlands program seems to be intrinsically geared to thinking about linear representations of Galois groups, and to set the scene, you have to begin by choosing a linear group, which will then cut everything else down in a Procrustean manner. At least superficially (and this answer reflects just superficial thoughts about the question), it doesn't seem well adapted to questions related to the nature of $\pi_1(\mathcal O)$, where no a priori linear structure is given, or indeed expected.