Butterworth low pass filter zeros location after bilinear transformation explanation
An $N^{th}$-order analog prototype system results after bilinear transform in an $N^{th}$-order discrete-time system. So the number of zeros and poles remains the same. All zeros at $|s|\rightarrow\infty$ map to $z=-1$. In your case there will be $6$ zeros at $z=-1$ because there are $6$ zeros at $|s|\rightarrow\infty$.

Of course, the magnitude (squared) response is symmetric, but the zeros at $|s|\rightarrow\infty$ cause the zeros in the magnitude at $\Omega\rightarrow\infty$ as well as at $\Omega\rightarrow -\infty$ (you can imagine those points meet at infinity). This becomes more obvious in the $z$-domain where moving towards positive frequencies means moving counterclockwise from $z=1$ (i.e., $\omega=0$), whereas moving towards negative frequencies means moving clockwise from $z=1$. In both cases you'll end up at $z=-1$.