I know that in general a curved space can have torsion or be torsion-free, however, can torsion exist in a flat space?
I'm guessing that it cannot for the reason that torsion is the antisymmetrization of the connection coefficients and they vanish in cartesian coordinates. Unless torsion, unlike the Riemann curvature, is not an intrinsic property of the space, but an artifact of the basis vectors changing independently along different paths.
The connection and the metric of a manifold are technically two independent quantities. If we allow both of those to be independent, then making a flat manifold with torsion is fairly trivial, for instance
\begin{eqnarray} ds^2 &=& -dt^2 + dx^idx_i\\ {\Gamma^\alpha}_{\mu\nu} &=& g^{\alpha\beta} \varepsilon_{\beta\mu\nu} \end{eqnarray}
This is one of the simplest connection with a torsion. it can easily be checked to have torsion, via
\begin{equation} {T^\alpha}_{\mu\nu} = {\Gamma^\alpha}_{[\mu\nu]} = g^{\alpha\beta} \varepsilon_{\beta\mu\nu} \end{equation}
with also (for what follows)
\begin{eqnarray} {T}_{\gamma\mu\nu} &=& g_{\alpha\gamma} g^{\alpha\beta} \varepsilon_{\beta\mu\nu}\\ &=& \varepsilon_{\gamma\mu\nu} \end{eqnarray}
and the contorsion tensor
\begin{eqnarray} K_{\alpha\beta\gamma} &=& \frac{1}{2} (T_{\alpha\beta\gamma} - T_{\beta\gamma\alpha} + T_{\gamma\alpha\beta})\\ &=& \frac{1}{2} T_{\alpha\beta\gamma} \end{eqnarray}
You can check fairly easily that despite the arbitrariness of our connection choice, it has torsion but it is still a metric connection :
\begin{eqnarray} \nabla_\alpha g_{\mu\nu} &=& \partial_\alpha g_{\mu\nu} - {\Gamma^\beta}_{\mu\alpha} g_{\beta\nu} - {\Gamma^\beta}_{\nu\alpha} g_{\mu\beta}\\ &=& - g^{\beta\gamma} \varepsilon_{\gamma\mu\alpha} g_{\beta\nu} - g^{\beta\gamma} \varepsilon_{\gamma\nu\alpha} g_{\mu\beta}\\ &=& - \varepsilon_{\nu\mu\alpha} - \varepsilon_{\mu\nu\alpha}\\ &=& 0 \end{eqnarray}
So that we have a connection that has torsion but is nethertheless metric, so that it still fits within the framework of the Einstein-Cartan theory, if we so wish. You can also check that this is flat via
\begin{eqnarray} {R^\alpha}_{\mu\nu\sigma} &=& 2 {\Gamma^\alpha}_{[\nu | \lambda} {\Gamma^\lambda}_{\sigma| \mu]}\\ &=& g^{\alpha\beta} g^{\lambda\beta} (\varepsilon_{\beta\nu\lambda} \varepsilon_{\gamma\sigma\mu} - \varepsilon_{\beta\mu\lambda} \varepsilon_{\gamma\sigma\nu})\\ &=& 0 \end{eqnarray}
As we are trying to contract a tensor symmetric in $\lambda \beta$ and one antisymmetric in it.
There is in fact a whole theory based on that notion called teleparallel gravity, which contains a flat connection with torsion, and it is an entirely acceptable theory of gravity.
