After checking the source code of the challenge, it was clear that the flag was used for two purposes: as AES key and as an IV.
AES is secure enough not to crack the key, but we can find out the IV with the following ‘trick’:
(the images at https://en.wikipedia.org/wiki/Block_cipher_mode_of_operation#CBC is always a good source of material to help thinking through what happens)
I replaced the first block of example encrypted text with the second block (so the blocks were in the following order now: c2|c2|c3), I kept the third block, so the padding remained correct and I decrypted it.
If we write down how the encrypted blocks created, we get this (p = plaintext, c = ciphertext, ^ = xor, E = AES encrypt, || = block boundary):
So if we want to know the IV (=FLAG) we can do this this way: IV = dec1 ^ p2 ^ c1 as we know all the ‘variables’.