Is using AES-256 with a key of less than 256 bits of entropy still quantum-resistant? Encrypting with AES-256 is widely claimed to be post-quantum secure.

But let's take a case where you use an initial key that has less than 256 bits of real entropy. In this example, let's work with a 128-bit key that has 128 bits of real entropy. We stretch it with something commonly used, like PBKDF2 with 120,000 iterations. We use the 256-bit result as the encryption key for AES-256 to encrypt some secret.

Is the the secret still secure against brute force attacks by a quantum computer, the kind can halve (or worse) the effective security of a symmetric cipher using Grover's algorithm? (Edited to add:) We assume the attacker knows the specifics of the above scheme, but not they length of the entropy or the initial key.
Currently, the quantum attacks work on the block cipher itself. The Grover's search algorithm reduces the complexity of 256-bit key into 128-bit since it has complexity $\mathcal{O}(\sqrt{n})$ and in the case of 256-bit $\mathcal{O}(\sqrt{2^{256}}) = \mathcal{O}(2^{128})$

Since you have started with 128-bit entropy, theoretically, the quantum attacker to block cipher itself with a 256-bit key will not benefit from Grover's algorithm since Grover's algorithm will search 256-bit. Keep in mind that 256-bit space is infeasible for the quantum. 

What if in an unforeseeable future, someone builds a quantum circuit for the PBKDF2 + AES combination that is the search for the 128-bit input to PBKDF2 and uses the output for the AES-256 key and perform Grover's kind of attack? 

This can be done by extending the AES-256 Grover Search with a 128-Q-bit PBKDF2 Quantum circuit. Since the input is 128-bit, the search will be $\mathcal{O}(2^{64})$ to find the ciphertext.

The attacker, always, will use the weakest point!.