Yes, you can convert the Hill Cipher into almost any mode of operation. For simplicity, assume that $M$ represent the key matrix and $P$ represent the plaintext vector and $C$ is the corresponding ciphertext vector with $C = M\cdot P \pmod{26}$. For simplicity, we omit $\bmod 26$ for the rest.
ECB mode: we can assume that Hill cipher is already in ECB mode. As usual, divide the message into the correct matrix size, you also need a padding scheme.
CBC mode: generate a random $IV$ which must be unpredictable, however, this is your least concern, as usual, $ C_0 = M\cdot(P_0\oplus IV) \text{ and } C_i = M\cdot(P_0\oplus C_{i-1})$. You also need a padding scheme.
CTR mode: $C=P \oplus [M \cdot (\text{nonce || Counter})]$. Here one may need to take care of the counter since it is a Matrix. One can consider each cell of the Matrix as a $\bmod 26$ counter. No need to padding.
CMC-MAC: Since you have CBC mode, you have it, now. Remember, in CMC-MAC the IV is all zero.
Note 1: For CBC-MAC we don't need just AES, we can use any secure block cipher.
Note 2: the $\oplus$ can generate a number greater than 26. You need to use $+ \bmod26$
And a final note: The Hill cipher is vulnerable to known-plaintext attack which means that it is broken! You cannot build a secure mode of operation from a broken cipher!