Probability of SHA256 Collisions for Certain Amount of Hashed Values
The wikipedia page for the Birthday problem gives the details, including the exact formula.

As an approximation you can use:

$$p \approx 1-\exp(-\frac{\tfrac{1}{2}n(n-1)}{2^{256}})
\approx 1-\exp\left(-\frac{n^2}{2 \cdot 2^{256}}\right)
\approx 1-\exp\left(-\tfrac{1}{2} \left(\frac{n}{2^{128}}\right)^2\right)$$

If $n \ll 2^{256/2}$ and thus $p \ll 1$ you can use the approximation $\exp(x) \approx 1+x$ and obtain:

$$p \approx \tfrac{1}{2} \left(\frac{n}{2^{128}}\right)^2$$

This means that the probability is negligible as long as you have significantly less that $2^{128}$ values. (For an idea of "how significantly" less, refer to the table on the Wikipedia page, "Desired probability of random collision".) That's the case for any realistic amount of data, so an unbroken 256 bit hash is good enough and there is no need to upgrade to 512 bits.