128-bit entropy simply means that we have $2^{128}$ different values to search, which is similar to 128-bit security. For a single target that is impossible since even the collaborative powers of Bitcoin Miners can reach $\approx 2^{92}$ in a year. Therefore one needs $2^{35}$ years to find the correct password. The iteration of PBKDF2 is 2048 so, we need to multiply $2^{35}$ with $2^{11}$ that makes $2^{46}$ years if we assume that SHA2d running time is equal to PBKDF2.
In the case of finding multi-password attack, with parallel versions of Oechslin's rainbow tables, the expected cost of finding a password from $t$ targets is $2^{128}/t$. Now assume that you have a billion of targets. Then, you will be able to find the first password much lower than 128-bit security. The cost would be below $2^{100}$ and the time would be below $2^{70}$. This can be achievable with supercomputers.
Conclusion: Using 128-bit entropy is not secure in multi-target attack case. The users should concentrate on password settings with 256-bit entropy.
Note 1: The title includes the word collision, however, a collision has nothing to do with finding a password of a user. A collision is finding two pairs of passwords such that after PBKDF2 is applied the result will be the same. This has cost $2^{64}$ with 50% probability of success. The pre-image attack is the attack for passwords with a cost of $2^{128}$ is the setup of the question.
Note 2: A detailed entropy calculations of Bip39 in this answer.