No, they meant -2^16, ie, -(0)

-2^16 is not -0. The unary minus binds more tightly than the binary xor, so this comes out as

  -2^16 = (-2) ^ 16 = 0b111...1110 ^ 0b10000 = 0b111...11011110 = -18

if I'm not mistaken.

No, we mean the power operator, 2 to the power of 16 would be '0' in a short int, since the value overflows...

But since philosophically a signed int has '1's in every bit from the MSB to infinity towards the left, it should thus be -0 not 0.

In two's complement (the usual binary representation of signed ints), -0 is the same as 0.