I'm going to take the step of disagreeing slightly with the perspective offered in Professor Emerton's answer, in a bold attempt to prise away the green for myself.

The question is, what does the Langlands program have to say about the finiteness (or otherwise) of $\pi_1(\mathcal{O}_K)$? Let's first revisit what one knows by ``classical'' methods. Minkowski proved that if $K$ is an extension of degree $n$, then

$$\Delta_K \ge \left(\frac{\pi}{4}\right)^{r_2} \frac{n^n}{n!},$$

which, asymptotically, implies that the root discriminant $\delta_K = |\Delta_K|^{1/n}$ is at least $e \sqrt{\pi/4} - \epsilon$ for large $n$, and in particular, that $|\Delta_K| > 1$ if $n > 1$. This is purely a geometry of numbers argument.

What does the Langlands program have to say about this question? First, let me ask a related question: what does the Langlands program say about elliptic curves $E/\mathbb{Q}$ with good reduction everywhere? In this case, it implies (and is known, by Wiles!) that there exists a classical cuspidal modular form $$f \in S_2(\Gamma_0(1)).$$ The latter group vanishes, and so $E$ does not exist. Yet this latter fact seemingly requires an actual computation

• namely, that $X_0(1)$ has genus zero. This could be tricky if one wants to replace $E$ by an abelian variety $A$ of (varying) dimension $g$, or replace $\mathbb{Q}$ by another field $F$. It turns out, however, that one can show that $f$ does not exist purely from the existence of the relevant functional equation - more on this later.

Let us return to our original question. Suppose we have a field $K$ unramified everywhere over $\mathbb{Q}$. It has been suggested that one should ponder the existence of algebraic automorphic forms $\pi$ for $\mathrm{GL}_n/\mathbb{Q}$ associated to irreducible Artin representations $\rho$ of $\mathrm{Gal}(K/\mathbb{Q})$. However, it is more natural to consider the regular representation. In this case, it is (of course) known by Hecke that $\zeta_K(s)$ has a meromorphic continuation that is entire away from $s = 1$, and, moreover, that $\zeta_K(s)$ satisfies a functional equation of a precise type. Now, purely given the existence of the functional equation for $\zeta_K(s)$, Odlyzko gives a formula(!) for $\log |\Delta_K|$ as some function of $s$ involving the roots of $\zeta_K$ (the function is (obviously) constant, but is not obviously constant). One then deduces a lower bound for $\delta_K$ of the form $2 \pi e^{\gamma} - \epsilon$, which is better than Minkowski's estimate. (Odlyzko's method can be refined to give better bounds.)

What is easy to miss in this argument is that the Langlands conjecture part of the problem - in this case the theorem of Hecke - is already known! One might claim that this argument uses "more" than Langlands and ask for an argument that is purely algebraic and geometric (and here by "one", I mean Brian Conrad or Chevalley), but I think this is a little misguided. After all, I think it would be hard to prove that there does not exist any Maass form for $\mathrm{SL}_2(\mathbb{Z})$ of eigenvalue $1/4$ without using some analysis.

Can one argue similarly to see that there are no abelian varieties $A$ with good reduction everywhere? Indeed, Mestre gave such an argument (before Fontaine!). Namely, if $A$ is an abelian variety, and $L(A,s)$ is automorphic in the expected sense, then $A$ has conductor at least $10^{g}$. Moreover, this is close to optimal ($X_0(11)^g$ has conductor $11^g$). There are other arguments along these lines. Stephen Miller (and Fermigier independently) proves that there are no cusp forms for $\mathrm{SL}_n(\mathbb{Z})$ of "weight zero" (cohomological with the same infinitesimal character as the trivial representation) and "level one" for all $n$ in the range $2 \ldots 23$ - another generalization of the fact that $X_0(1)$ has genus zero - using Rankin--Selberg L-functions.

Here is a final reason why one should expect some analysis. All of these arguments fundamentally require the discriminant of $K$ to be small. Any ``algebraic'' method should be expected to work more generally. Yet consider the question of whether $\pi_1(\mathcal{O}_K)$ is finite when $\delta_K$ is large. The only known method for answering this question is producing an unramified extension $L/K$ such that the GS-inequality applies to $\pi_1(\mathcal{O}_L)$. The question on whether these groups are (always?) infinite when $\delta_K$ is sufficiently large is wildly open, and I know no good heuristics on this question.