At what point does passphrase strength match the strength of the cipher?
If I'm not missing some important point here, that's rather trivial to calculate. As implicitly stated in your question, the attacker has two possible ways to brute-force the encryption key: either by brute-forcing the encryption key itself or by brute-forcing the user-entered password.

Given the numbers of iterations of the key derivation function n, the time per kdf iteration tkdf, the length of the encryption key b (in bits) and the time per invocation of the actual encryption algorithm tenc, the average time needed to crack the encryption key itself (by testing half of the available keyspace) is

1/2&sdot;2b&sdot;tenc.

The average time to crack the user-entered password is

1/2&sdot;K&sdot;(n&sdot;tkdf+tenc),

with K=cl being the entirety of all passwords available to the user (c: number of characters available, l: length of password).

In your scenario, tenc=tkdf=1, and n=327661. Assuming AES-256 is used (b=256), one has to solve the equation

1/2&sdot;2b&sdot;tenc = 2256 = K&sdot;327662 = 1/2&sdot;K&sdot;(n&sdot;tkdf+tenc),

which yields (approximately) K=5.9447&sdot;1035. Assuming an available keyspace of 90 characters (which most users aren't likely to even find on their keyboards), this translates to approximately l=36.

(Disclaimer: I'm neither a mathematician nor an crypto expert. In case of any mathematical or semantic error, please comment.)

Related: https://security.stackexchange.com/questions/74566/long-passwords-and-key-derivation-functions