n-bit majority digital logic
I think you can reduce this (for odd n) to a set of AND gates followed by an OR gate. 

For n voters you have a majority if (n+1)/2 agree. So this is an n pick k situation where k = (n+1)/2 and you need 

z = \$\frac{n!}{k!(n-k)!}\$ k-input AND gates and a z-input OR gate (where n is odd and k = \$\frac{n+1}{2}\$) 

For even n it would be similar except a majority is n/2 +1. 



For the n choose k algorithmically, I refer you to this Python code.