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.
[Added July 14, in response to Minhyong's question in the comments as to whether or not discrete series can convert into non-discrete series after applying some functoriality.
The answer is essentially no, as I will now explain.
Added November 29, 2011: What follows is wrong; the answer seems rather to be yes.
(See below for details.)
For an arithmetic geometer, one should think of an automorphic form on the adelic group
$G(\mathbb A_K)$ as a morphism from the motivic Galois group (over the base number field $K$) to the $L$-group of $G$. (There are subtleties and caveats, of course, but they need not concern us here; all I will say about them is that automorphic forms can give rise to "motives" with non-integral $(p,q)$ in its Hodge decomposition, which necessitates enlarging the category of motives to an unknown larger category, whose hypothetical Tannakian group is called "the Langlands group".)
Now functoriality takes place when you have a map from the $L$-group of $G$ to the $L$-group of $H$; one can just compose this with a map from the motivic Galois group to the former, to obtain a map from the motivic Galois group to the latter. Functoriality is the assertion
that the corresponding automorphic form on $H(\mathbb A_K)$ exists.
Now given an automorphic form $\pi$, its factors at the primes $v|\infty$ encode (via the local Langlands corresondence for $\mathbb R$ or $\mathbb C$) the Hodge numbers of the corresponding motive. One feature of discrete series is that (among other properties) they give rise to regular Hodge numbers, i.e. to sequences of $h^{p,q}$ with each $h^{p,q}
\leq 1$. Now our original automorphic rep'n $\pi$ on $G(\mathbb A)$ corresponds to a motive whose Mumford--Tate group lies in the $L$-group of $G$, and if $\pi$ has discrete series components at primes above $\infty$, it has regular Hodge numbers at every place dividing $\infty$. If we then pass to a new motive by applying some map from the $L$-group of $G$ to the $L$-group of $H$, then concretely this corresponds to doing some kind of multilinear algebra on our motive, and the only way this can kill the property of having regular Hodge--Tate weights is if we do something like taking the diagonal map from the $L$-group of $G$ into its product with itself, and then embed the latter into the $L$-group of $H$.
All such constructions will necessarily be a "reducible" rep'n of the $L$-group of $G$
in the $L$-group of $H$ (more precisely, the centralizer will be a non-trivial Levi),
and the corresponding automorphic form won't be a cuspform.
But even if we destroy the property of having regular Hodge numbers, we typically still
don't have an Artin motive. To get an Artin motive we have to have $h^{p,q} = 0$ unless
$p = q = 0$, and to do this, we have to do even more destructive things, like
map the $L$-group of $G$ into the $L$-group of $H$ via the trivial representation,
or something similar.
Again, this won't correspond to any kind of interesting automorphic forms, just
those that correspond to (certain) sums of characters. So we can't produce interesting
Galois type automorphic forms out of automorphic forms whose factors at primes above
$\infty$ are discrete series.]
[Correction added Nov. 29, 2011: From the Galois/motivic point of view, we have an algebraic group (the Mumford--Tate group of some motive), with a representation (the particular motive), and the Mumford--Tate group contains a cocharacter whose eigenvalues are the Hodge numbers. Discrete series corresponds to the eigenspaces being one dimensional.
We now apply some functoriality, which is essentially to say that we apply some multi-linear algebraic process to the representation. Now this can certainly produce eigenspaces for the cocharacter of multiplicity $> 1$. (E.g. the adjoint representation of $SL_3$ has a two-dimensional eigenspace.) So it seems that functoriality doesn't preserve being discrete series. It does preserve being tempered. And the remarks about not getting Artin motives still seem okay, since while the eigenspaces can become greater than $1$-dimensional, for all the eigenspaces to become trivial, we have to do something pretty destructive, like applying functoriality for the trivial representation.]
Best Answer
The point is that it is one thing to show that two mathematical objects are isomorphic; it is another (stronger) thing to give a particular isomorphism between them. A rather concrete instance of this is in combinatorics, where if $(A_n)$ and $(B_n)$ are two families of finite sets, one could show that $\# A_n = \# B_n$ by finding formulas for both sides and showing they are equal, but it is preferred to find an actual family of bijections $f_n: A_n \rightarrow B_n$.
This is not just a matter of fastidiousness or a general belief that constructive proofs are better. When considering functorialities between various isomorphic objects, the choice of isomorphism matters. For instance, often one wants to put various isomorphic objects into a diagram and know that the diagram commutes: this of course depends on the choice of isomorphism.
In the case of class field theory, these functorialities take the form of maps between the abelianized Galois groups / norm cokernel groups / idele class groups of different fields. The isomorphisms of class field theory can be shown to be the unique ones which satisfy various functoriality properties (and some "normalizations" involving Frobenius elements), and this uniqueness is often just as useful in the applications of CFT as the existence statements.
All of this, by the way, is explained quite explicitly in Milne's (excellent) notes: you just have to read a bit further. See for instance Theorem 1.1 on page 20: "There exists a unique homomorphism...with the following properties [involving Frobenius automorphisms and functoriality]..."
As a final remark: it is important to note that the word "canonical" in mathematics does not have a canonical meaning. To say that two objects are canonically isomorphic requires further explanation (as e.g. in the Theorem I mentioned above). Even the "unique isomorphisms" that one gets from universal mapping properties are not unique full-stop [generally!]; they are the unique isomorphisms satisfying some particular property.