As many have already observed you will have a small increase in the size of your file after each level of encryption, due to the IV that needs to be added to the file after each encryption. Not really relevant.
Rather, I'd like to observe that your motivation for doing that is obviously to increase the robustness of your ciphertext against attacks, including brute-force ones. If you use a key of $n$ bits for each of $h$ levels of encryption, let's say keys $k_1, k_2, \ldots, k_h$, you maybe expect to get the robustness of one unique encryption based on one longer key of $h\times n$ bits. Theoretically speaking, it is possible to run the Meet-in-the-Middle attack, that allows an adversary to reduce the size of the key space to less than $2^{h\times n}$. A practical example is 2-DES, where the plaintext $P$ is first encrypted by a DES key (56 bits), thus obtaining a ciphertext $C'$, and then $C'$ is again encrypted by another DES key, thus obtaing the final ciphertext $C$. However, the expectation of having a size of keyspace equal to $2^112$ is wrong. The actual size will be $2^57$ because the Meet-in-the-Middle attack, that is a known-plaintext attack (meaning that the adversary knows a pair $(P, C)$), will allow the adversary to first build a table of all $2^56$ possible encryptions of $P$ (one for each potential key) and then generate all the possible decryptions of $C$ (again, one for each potential second key) and, for each of them, let's call it $C''$, check whether $C''$ is equal to some of the potential ciphertexts in the table. In case of match we got $C'' = C'$ and we have discovered the two DES keys. Total number of encryptions/decryptions will be $2^56\times 2 = 2^57$.
For similar reasons, 3-DES (three levels of DES encryption using three different keys) offers the security of a 112 bits key.
