Is there a maximum amount of data which can be encrypted using a symmetric (AES256) key? I am working on an application where we will be encrypting billions of small (&lt; 2,000 bytes) records.  Combined, we are facing many TB of data.  Each record will be encrypted independently, using a unique IV/nonce.

We are aware of the restrictions around the reuse of IV/nonce values and the limits on the amount of data which can be encrypted using a key/IV pair (the NIST SP 800-38 series).  Given that each record will use a unique IV, this will likely be the primary factor driving the rotation of keys.

Is there a maximum amount of data - regardless of the number of encryption operations using a unique IV/nonce - where the AES key itself should be rotated, or is the retirement driven entirely by the restrictions set by block chaining mode?
You should generally never use a single AES key for more than a few petabytes of data because of the so-called birthday bound: Cryptosystems based on AES like AES-GCM tend to have distinguishing advantages bounded only by $$\frac{q(q - 1)/2}{2^b}$$ after processing $q$ blocks of data, where $b = 128$ is the block size, arising from the birthday paradox.

When $q \approx \sqrt{2^b} = 2^{64}$, the standard theorems provide essentially no security guarantees.  See sweet32 for what can go wrong when you reach the birthday bound of a block cipher—of course the birthday bound is much smaller for a 64-bit block cipher like Blowfish, where it is $2^{32}$, but the idea is still relevant to AES.

(Had we standardized a 256-bit block cipher like Rijndael-256 or Threefish, the birthday bound of $2^{128}$ would be completely irrelevant.  But we didn't, and while there are 256-bit block ciphers out there, hardly anyone ever uses them.)

Particular cryptosystems using AES may have smaller limits—for example, AES-GCM with a random nonce is unsafe for more than a few billion messages under the same key, because the nonce is only 96 bits long and the security degrades catastrophically in case of collision, which is expected to happen after only about $\sqrt{2^{96}} = 2^{48}$ messages.