The "degree" of a line bundle does not make sense in general, it is better to study $\textrm{Pic}^0$, the connected component of the indentity of the Picard scheme. For abelian varieties this is exactly the dual abelian variety.

As noted in the comments, the Picard group of a linear algebraic group $G$ is finite, so $\textrm{Pic}^0(G)=0$ and hence it is not very interesting. By Chevalley's theorem any algebraic group is an extension of an abelian variety by a linear algebraic group (as already noted by P Vanchinathan). Hence as a *variety* it is a product $G \times A$ where $G$ is linear algebraic and $A$ is an abelian variety. Moreover we have $\textrm{Pic}^0(G \times A) = \textrm{Pic}^0(G) \times \textrm{Pic}^0(A) = \widehat{A}$, so one can reduce to the case of abelian varieties.