The answer is $\boxed{1}$.

All terms are even. Therefore, in order for any one to be a square, it must be congruent to $0$ or $4$ modulo $16$. Only the first term satisfies this requirement. And indeed $4$ is a square. (Note that the remainder when being divided by $16$ is determined by its last three digits. So we only need to check the first three terms.)

This problem can also be reasoned using the conclusion of # 4147. Because $4$ is a square number, therefore dividing every term of this sequence by $4$ will not change the fact whether or not that term is a square. By # 4147, we know there is only one term in $$1, 11, 111, 1111,\cdots$$

is square. So there is only one square in the original sequence as well.