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.

The "degree" of a line bundle is not defined 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, we may write any algebraic group $H$ as $$0 \to G \to H \to A \to 0,$$ where $G$ is linear algebraic and $A$ is abelian. I claim that $\textrm{Pic}^0(H) = \textrm{Pic}^0(A) = \widehat{A}$, so one can reduce to the case of abelian varieties. Here is a sketch of a proof. By [1, Prop. 6.10], we have an exact sequence $$0 \to \textrm{Pic}(A) \to \textrm{Pic}(H) \to \textrm{Pic}(G).$$ As already noted, $\textrm{Pic}(G)$ is finite. Hence when we restrict this exact sequence, to the identity component, we obtain $\textrm{Pic}^0(H) = \textrm{Pic}^0(A)$, as required.

[1] Sansuc - Groupe de Brauer et arithmétique des groupes algébriques linéaires.