I came across a comment about AES 128 without ShiftRow. I want to use this weakness to perform an attack. So far, I get that It possible to divide in 4 independent blocks of 32 bits.
Now, brute force and differential/linear based attacks are not likely to work. However, there is a different approach.
It is effectively 4 different 32 bit block ciphers, interleaved.
So, how would you break a 32 bit block cipher? By an attack on the limited size of the code book.
If the person using the modified AES cipher was silly enough to use CTR mode, it's easy - an increment of the counter almost never modifies 3 of the 4 32-bit block ciphers, hence 3/4 of the plaintext is easy to deduce. The last 1/4 might be harder, however at that point, I would claim it is, by any reasonable standards, broken.
If we consider CBC mode (where the inputs to the block cipher is effectively randomized by xor'ing in the previous ciphertext block), we can consider what happens if you have, say, a Megabyte of ciphertext output. At that point, you have 256K inputs for each of the 4 32-bit block ciphers, that is, 64k block inputs. At this level, we would expect several collisions (that is, places where two inputs to the same block cipher are the same, resulting in the same output); each collision would reveal the xor of two 4-byte sections of the plaintext. While not nearly as bad as the CTR mode case, it can still reveal more information than we'd like (and the probability that any section of the plaintext is involved in a leak increases with the amount of ciphertext that is available to the attacker).