How many 12-bit strings with more 1’s than 0’s?
Your expression is correct. Here is another way that exploits the symmetry.
Let $a$ be the number of strings with more $1$'s than $0$'s, let $b$ be the number with more $0$'s than $1$'s, and let $c$ be the number with equal numbers of $0$'s and $1$'s.
Then $a=b$, $a+b+c=2^{12}$, and $c=\binom{12}{6}$. So $2a=2^{12}-\binom{12}{6}$ and therefore $$a=2^{11}-\frac{1}{2}\binom{12}{6}.$$