Here's some piece of the bigger picture. Maass forms and holomorphic modular forms are both automorphic representations for $GL(2)$ over the rationals. An automorphic representation is a typically huge representation $\pi$ of an adele group (in this case $GL(2,\mathbf{A})$, with $\mathbf{A}$ the adeles of $\mathbf{Q}$). Because the adeles is the product of the finite adeles and the infinite adeles, this representation $\pi$ is a product of a finite part $\pi_f$ and an infinite part $\pi_\infty$. The infinite part is a representation of $GL(2,\mathbf{R})$ (loosely speaking -- there are technicalities but they would only cloud the water here).
The representation theory of $GL(2,\mathbf{R})$, in this context, is completely understood. The representations basically fall into four categories, which I'll name (up to twist):
1) finite-dimensional representations (these never show up in the representations attached to cusp forms).
2) Discrete series representations $D_k$, $k\geq2$ (these are the modular forms of weight 2 or more).
3) The limit of discrete series representation $D_1$ (these are the weight 1 forms).
4) The principal series representations (these are the Maass forms).
Now what does Langlands conjecture? He makes a conjecture which does not care which case you're in! He conjectures the existence of a "Galois representation" attached to $\pi$, and this is a "Galois representation" in a very loose sense: it is a continuous 2-dimensional complex representation of the conjectural "Langlands group", attached to $\pi$. Note that there should be a map from the Langlands group to the Galois group, and in the case of Maass forms and weight 1 forms Langlands' representation should factor through the Galois group. For modular forms of weight 2 or more Langlands' conjecture has not been proved and in some sense it is almost not meaningful to try to prove it because no-one can define the group. In particular Deligne did not prove Langlands' conjecture, he proved something else.
So Clozel came along in 1990 and tried to put Deligne's work into some context and he came up with the following: he formulated the notion of what it meant for $\pi_\infty$ to be algebraic (in fact there are two notions of algebraic, which differ by a twist in this context, so let me write "$L$-algebraic" to make it clear which one I'm talking about) and conjectured that if $\pi$ were $L$-algebraic then there should be an $\ell$-adic Galois representation $\rho_\pi$ attached to $\pi$. Maass forms with eigenvalue $1/4$, and holomorphic eigenforms, are $L$-algebraic, and the $\ell$-adic Galois representation attached to the Maass forms/weight 1 forms is just the one you obtain by fixing an isomorphism $\mathbf{C}=\overline{\mathbf{Q}}_\ell$. I should say that Clozel worked with $GL(n)$ not $GL(2)$ and also worked over an arbitrary number field base.
Whether or not the image of $\rho_\pi$ is finite is something which is conjecturally determined by $\pi_\infty$: you can read it off from the infinitesimal character of $\pi_\infty$ and also from the local Weil group representation attached to $\pi_\infty$ by the local Langlands conjectures, which are all theorems (of Langlands) for real reductive groups.
Put within this context your question becomes purely local: one has to figure out what Clozel's recipe gives in each case to get a handle on what your question is asking. You're asking about principal series representations. If you work out Clozel's recipe in these cases you find that if $\lambda\not=1/4$ then $\pi_\infty$ is not $L$-algebraic (and so we don't even expect a representation of the Galois group, we just expect a representation of the conjectural Langlands group), and if $\lambda=1/4$ then, up to twist, we expect the image to be always finite, because, well, that's what the calculation gives us.
I learnt this by just doing all these calculations myself. I wrote them up in brief notes at http://www2.imperial.ac.uk/~buzzard/maths/research/notes/automorphic_forms_for_gl2_over_Q.pdf and http://www2.imperial.ac.uk/~buzzard/maths/research/notes/local_langlands_for_gl2R.pdf (both available from http://www2.imperial.ac.uk/~buzzard/maths/research/notes/index.html ).
So why is there this asymmetry? Well actually this asymmetry is not surprising because it is predicted on the Galois side as well. If you look at an irreducible mod $p$ ordinary Galois representation which is odd then its universal ordinary deformation is often known to be isomorphic to a Hecke algebra of the type defined by Hida (so in particular we get lots of interesting $\ell$-adic Galois representations with infinite image). In particular its Krull dimension should be 2 (and this was already known to Mazur in the 80s). But the calculations for these Krull dimensions involve local terms, and the local term at infinity depends on whether the representation is odd or even. If you consider deformations of an even Galois representation then the calculations come out differently and the Krull dimension comes out one smaller. In particular one only expects to see finite image lifts, plus twists of such lifts by powers of the cyclotomic character.
So in summary you see differences on both sides -- the automorphic side and the Galois side -- and they match up perfectly! You don't expect $\ell$-adic representations to show up in the Maass form story and yet things are completely consistent anyway.
Toby Gee and I recently tried to figure out the complete conjectural picture about how automorphic representations and Galois representations were related. Our conclusions are at http://www2.imperial.ac.uk/~buzzard/maths/research/papers/bgagsvn.pdf . But for $GL(n)$ this was all due to Clozel over 20 years ago (who would have known all those calculations that I linked to earler; these are all standard).
Best Answer
Here's an answer that's more work than Kevin Ventullo's, but works for $\ell=2$ and $\ell=3$, and gives you a method that will work in various more general cases.
There is a surjection $\pi_1\colon G_\mathbb Q \to \overline{\rho}_{f,l} (G_{\mathbb Q})$ and a surjection $\pi_2 \colon G_\mathbb Q \to \operatorname{Gal}(\mathbb Q(\zeta_\ell) / \mathbb Q)$. The group $\overline{\rho}_{f,l} (G_{\mathbb Q(\zeta_\ell)})$ is the image under $\pi_1$ of the kernel of $\pi_2$. In other words, it is the intersection of the image of the product map $(\pi_1,\pi_2) \colon G_\mathbb Q \to \overline{\rho}_{f,l} (G_{\mathbb Q}) \times \operatorname{Gal}(\mathbb Q(\zeta_\ell) / \mathbb Q)$ with $G_{\mathbb Q}) \times 1$.
By Goursat's lemma, the image of the product map is $ \{ (a,b) \mid f_1(a) =f_2 (b) \}$ where $f_1 \colon \overline{\rho}_{f,l} (G_{\mathbb Q}) \to Q$ and $f_2 \colon \operatorname{Gal}(\mathbb Q(\zeta_\ell) / \mathbb Q) \to Q$ are surjections for some finite group $Q$. Expressed in this language, $\overline{\rho}_{f,l} (G_{\mathbb Q(\zeta_\ell)})$ is the kernel of $f_1$.
Now because $f_2$ is surjective, $Q$ is a quotient of a cyclic group of order $\ell-1$, and thus is a cyclic group of order dividing $\ell-1$.
So $\overline{\rho}_{f,l} (G_{\mathbb Q(\zeta_\ell)})$ is the kernel of a map from $\overline{\rho}_{f,l} (G_{\mathbb Q(\zeta_\ell)})$ to a cyclic group of order dividing $\ell-1$. Thus it contains the kernel of a map from $SL_2(\mathbb F_\ell)$ to the kernel of a cyclic group of order dividing $\ell-1$.
Since the abelianization of $SL_2(\mathbb F_\ell)$ is trivial (for $\ell>3$) or has order relatively prime to $\ell-1$ (for $\ell=2,3$, since $2$ is relatively prime to $1$ and $3$ is relatively prime to $2$), $\overline{\rho}_{f,l} (G_{\mathbb Q(\zeta_\ell)})$ contains $SL_2(\mathbb F_\ell)$.
Since $SL_2(\mathbb F_\ell$ acts absolutely irreducibly on the standard representation for all $\ell$, it follows that $\overline{\rho}_{f,l} (G_{\mathbb Q(\zeta_\ell)})$ acts absolutely irreducibly on the standard representation, as desired.
It immediately follows that the image of $\operatorname{Sym}^n \overline{\rho}_{f,\ell} \mid_{ G_{\mathbb Q(\zeta_\ell)}}$ containsthe image of $SL_2(\mathbb F_\ell)$ under the $n$th symmetric power representation. Since the $n$th symmetric power representation of $SL_2(\mathbb F_\ell)$ is absolutely irreducible for $n \leq \ell-1$, the representation $\operatorname{Sym}^n \overline{\rho}_{f,\ell} \mid_{ G_{\mathbb Q(\zeta_\ell)}}$ is absolutely irreducible for $n\leq \ell-1$.