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 $D(D-3)/2$.
