In string theory, the first excited level of the bosonic string can be decomposed into irreducible representations of the transverse rotation group, $SO(D-2)$. We then claim that the symmetric traceless part (i.e. the 35 rep) is the spin-2 graviton – but isn't the label "spin-2" intrinsically 3+1 dimensional? I.e. it labels the representation under the little group $SU(2)$?
String Theory – Why Does the Graviton Have Spin-2?
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The spin $s$ of a particle characterizes how the rotation generators act on it. In $D$ dimensions, you represent the little group $SO(D-1)$ for massive particles and $SO(D-2)$ for massless ones. In fact, you really need to consider its universal cover $\textrm{Spin}(n)$ which happen to be just its double cover.
Now, you can define the spin to be the largest real $s$ such that $$ \mathrm{exp}\left(\frac{2i\pi}{s}J\right) = \rm{id} $$ where $J$ is any generator in your representation. This is basically the statement that, to return to identity, you need to do a $4\pi$ rotation for spin $\frac{1}{2}$, a $2\pi$ for spin 1, a $\pi$ for spin 2... Note that because the universal cover is the double cover, $s$ has to be a half-integer.
It is clear that Lorentz vectors have spin 1. Let's take a symmetric 2-tensor $T_{\mu\nu}$. It transforms as $T'_{\alpha\beta} = R^{\:\,\mu}_{\alpha}R^{\:\,\nu}_{\beta}T_{\mu\nu}$, where $R = \mathrm{exp}(i\theta J)$ are the usual $SO(n)$ matrices. Infinitesimally, \begin{align} \delta T_{\alpha\beta} & = i\theta(J^{\:\,\mu}_{\alpha}\delta^{\:\,\nu}_\beta + J^{\:\,\nu}_{\beta}\delta^{\:\,\mu}_\alpha)T_{\mu\nu} \\ & = 2 i\theta J^{\:\,\mu}_{\alpha}T_{\mu\beta} \\ \end{align} where the symmetry of the tensor is crucial. Now, because $\mathrm{exp}(2i\pi J) = 1$, you see that for $\theta = \pi$, you get back to identity. This is spin 2!
You can generalize this to show that traceless symmetric n-tensors are spin n. You need them to be traceless because you want irreducible representations. With this, it shouldn't be too hard to derive the degrees of freedom of a spin $s$ particle in $D$ dimensions e.g. for the graviton, it is $D(D-3)/2$.
Let $\phi : \mathbb{R}^4\to V$ be a field with (complex) target vector space $V$, transforming in a finite-dimensional projective representation $\rho_\text{fin} : \mathrm{SO}(1,3)\to\mathrm{U}(V)$. As it is a field, the representation of the translations $\mathbb{R}^4$ on $V$ is the trivial one, since the field transforms as $\phi(x)\overset{x\mapsto x+a}{\mapsto} \phi(x+a)$. Hence, the field transforms in a finite-dimensional representation $\sigma_\text{fin}$ of the Poincaré group, but the non-trivial, i.e. interesting, part is the representation of the Lorentz group. Hence, your premise that we "only study finite-dimensional representations of the Lorentz group" is wrong, it's just that the finite-dimensional translations are always represented by their trivial representation.
In the quantum field theory, the field now becomes operator-valued, acting upon some Hilbert space $\mathcal{H}$. Since the quantum field theory shall have Poincare symmetry, there must be a projective unitary representation $\sigma_\text{U} : \mathrm{SO}(1,3)\ltimes\mathbb{R}^4\to\mathrm{U}(\mathcal{H})$ upon this space of state. By one of the Wightman axioms, we have that $$ \sigma_\text{fin}(\Lambda,a)\phi(\Lambda^{-1} x-a) = \sigma_\text{U}(\Lambda,a)^\dagger \phi(x)\sigma_\text{U}(\Lambda,a)\quad \forall \Lambda\in\mathrm{SO}(1,3),a\in\mathbb{R}^4$$ where on the l.h.s., $\sigma_\text{fin}$ is a finite-dimensional matrix acting upon the vector $(\phi^1,\dots,\phi^{\dim(V)})$, and on the r.h.s., the $\sigma_\text{U}$ are operators on $\mathcal{H}$ are multiplied with each component operator $\phi^i$.
We study the finite-dimensional representations because of this relationship - we have to know the finite-dimensional representations to be able to give the "classical" field, and we have to know the infinite-dimensional unitary representation to know how the Poincaré symmetry acts on states, and because the irreducible unitary representations correspond to particles by Wigner's classification. Since the Poincaré group is just as non-compact as the Lorentz group, these are also all infinite-dimensional.
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It is traditional to label massless (and some massive) states in higher dimension by their 3-d "spin" counterparts, even thought he label is completely inaccurate, as you say. All antisymmetric forms are "spin-1", the symmetric two-index object is "spin-2", a fundamental spinor is "spin 1/2" and a vector of spinors is "spin 3/2". These labels refer to the maximum helicity of the associated massless particle, although the number of components is completely different than in 4d. For learning the higher dimensional rotation group, there is an article by Scherk from the 1970s.