This is a problem that revolves around symmetry. I recognize that if there is a 4-bit string that it will have 1110 as an answer, but it will also have 0111 as an answer. The thing is, I'm not sure how to use this to get an answer.
I know there are $2^n$ different strings total. Would the fact that for each string that fits the constraints, there is a complement of that string that does not work (ie: 1110 works, but the complement 0001 does not) mean that half of the strings do not work? If so I believe the answer will be: $(2^n)/2$ or $2^{n-1}$ but I'm not sure if this is a fair argument.
You are not quite correct. It is true that there is a symmetry between strings with $k\ 1$s and strings with $n-k\ 1$s. If $n$ is odd, half the strings have less than $\frac n2\ 1$s. If $n$ is even, you need to account for the ones that have exactly $\frac n2\ 1$s, of which there are ${n \choose n/2}$, so for $n$ even the count is $\frac 12\left(2^n-{n \choose n/2}\right)+{n \choose n/2}=2^{n-1}+\frac 12{n \choose n/2}$