Are connections in general relativity spin-3 objects? If we see metric-compatible connections (not Levi-Civita) then can we say that the connections are spin-3 objects? If not, when do we say that an object with $J$ index is a spin-$J$ object ($J \in \mathbb{Z}$) ?
 A: Spin and rank
(For simplicity I will discuss only integer spin fields)
It is not precise to identify the rank of the field to the spin of the field.
For instance, a spin $s=0$ field is usually best described by a scalar field.
$$\phi(x)$$ But we can form tensors of higher rank by taking derivatives of the original field
$$B_\mu = \partial_\mu \phi(x)$$
Here I already know that the physical content of my vector field $B_\mu$ is a spin $s=0$ field.
In general, one needs more equations to specify the spin of the field. Weinberg's Qauntum Theory of Fields ch5. does a wonderful job at explaining the bottom up approach - if we want to describe a field of some spin $s$, let's see into which objects we can pack it up, and see what conditions these objects must satisfy.
Once the candidates are tried out, one can infer that, if you want massive relativistic fields of definite integer spin they must satisfy the following equations (the dots indicate that any rank tensor is possible):
\begin{align}
(\Box + m^2)\phi^{\mu\nu...} = 0\\
\eta_{\mu\nu}\phi^{\mu\nu....}=0\\
\partial_\mu\phi^{\mu...}=0\\
\phi^{\mu\nu...} - \text{ Totally symmetric in all indices}
\end{align}
(On-shell, completely symmetric, traceless and transverse)
Then, if the previous relations are valid, the rank of the tensor corresponds to the spin.
For massive fields of any integer spin, this was figured out by Fierz and Pauli, for massless field of any spin by Fronsdal. The massless case is more subtle, it turns out that if we want Lorentz covariance, we neeed our fields to also be gauge invariant.
Spin 2
The reason we say general relativity is described by a spin two field is because when we linearise the theory around flat space, the perturbation field $h_{\mu\nu}$ is a spin two field
$$g_{\mu\nu} = \eta_{\mu\nu} + h_{\mu\nu}$$
Now, in this linearised view, the connection would be
$$\Gamma_{\mu \nu}^{\sigma}=\frac{1}{2} \eta^{\sigma \rho}\left(\partial_{\nu} h_{\rho \mu}+\partial_{\mu} h_{\rho \nu}-\partial_{\rho} h_{\mu \nu}\right)$$
and it still describes a spin $s=2$ field, for the same reason above vector field describes a scalar field.
