Question: If I specify a metric tensor in terms of a choice of bases in the co-tangent space at a given point in a manifold, how much information does the metric tell me about the curvature tensor at that point? Can I calculate the curvature tensor entirely from the metric tensor?

Answer: 
Yes you can, but you need to know the metric up to the second derivative, so the metric in a neighb. of the manifold. You can do that for the Levi-Civita connection, since it is completely defined by the metric $g$.
Of course. Given the components of the metric tensor $ g\_ {ij} $ you can calculate the Christoffel symbols
$$\Gamma^i\_{kl}=\frac{1}{2}g^{im} \left(\frac{\partial g\_{mk}}{\partial x^l} + \frac{\partial g\_{ml}}{\partial x^k} - \frac{\partial g\_{kl}}{\partial x^m} \right)=\frac{1}{2}g^{im} \frac{\partial g\_{mk}}{\partial x^l} + \frac{1}{2}g^{im} \frac{\partial g\_{ml}}{\partial x^k} - \frac{1}{2}g^{im} \frac{\partial g\_{kl}}{\partial x^m},$$
 then you can calculate the components of the curvature tensor $ R\_ {ijkl} $
$$R\_{iklm}=\frac{1}{2}\left( \frac{\partial^2g\_{im}}{\partial x^k \partial x^l} + \frac{\partial^2g\_{kl}}{\partial x^i \partial x^m} - \frac{\partial^2g\_{il}}{\partial x^k \partial x^m} - \frac{\partial^2g\_{km}}{\partial x^i \partial x^l} \right) +g\_{np} \left( \Gamma^n\_{kl} \Gamma^p\_{im} - \Gamma^n\_{km} \Gamma^p\_{il} \right).$$
 It is a very long procedure, but it is possible.n