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.

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. When $X$ is a smooth projective curve, divisors mod algebraic equivalence is just $\mathbb{Z}.$