Even if we might not be able to speak of degree without reference to a given ample divisor, we can still speak of codimension-1 cycles being algebraically or rationally equivalent to 0 on a smooth projective variety $X$ of any positive dimension.  In this context, ${\rm Pic}^{0}(X)$ is the group of codimension-1 cycles algebraically equivalent to 0 modulo the subgroup of those rationally equivalent to 0. 

EDIT:  As suggested by the above, ${\rm Pic}^{0}(X)$ is the kernel of the map from divisors mod rational equivalence to divisors mod algebraic equivalence.