Claim : if $k$ is perfect of characteristic $0$, then :

$$\mathbb Z^{k, alg}\simeq \mathbb G_a\times D_k(k^*)$$

Here $\mathbb Z^{k, alg}$ stands for the pro-algebraic completion of $\mathbb Z$ (the group you are interested in), $\mathbb G_a$ is the additive group, and $D_k(M)$ denotes as usual the diagonalisable $k$-group associated to the abstract abelian group $M$.

Sketch of a proof.

We denote $G=\mathbb Z^{k, alg}$ to simplify. By definition, this is the Tannaka group of the Tannaka category ${\rm Rep}_k \mathbb Z$, that is canonically isomorphic to the category ${\rm Iso_k}$ whose objects are couples $(V,\phi)$, where $V$ is a finite dimensional $k$-vector space, and $\phi\in {\rm GL}(V)$.

Since $\mathbb Z$ is abelian, so is $G=\mathbb Z^{k, alg}$, hence since $k$ is perfect, it splits canonically into a product diagonalizable x unipotent :
$G=G^m\times G^u$.

Moreover $G^m=D_k(\Gamma)$, where
$\Gamma={\rm Hom}_{gr}(G,\mathbb G_m)={\rm Pic}({\rm Rep}_k G)\simeq {\rm Pic}({\rm Iso}_k)\simeq k^*$.

On the other hand $G^u$ is the tannaka group of the category $({\rm Rep}_k G)^u\simeq({\rm Iso}_k)^u={\rm Uni}_k$, where ${\rm Uni}_k$ is the category whose objects are couples $(V,\phi)$, where $V$ is a finite dimensional $k$-vector space, and $\phi$ is a unipotent endomorphism.

Now since $k$ is of characteristic $0$, there is an equivalence
${\rm Rep}_k(\mathbb G_a)\simeq {\rm Uni}_k $ sending $\rho :\mathbb G_a\rightarrow {\rm GL}(V)$ to $(V,\rho(1))$, compatible with fiber functors, so one deduces $G^u\simeq \mathbb G_a$.

Complement, probably related to your original interest.

All this is implicitely contained in

Saavedra Rivano, Neantro

Catégories Tannakiennes.

Lecture Notes in Mathematics, Vol. 265. Springer-Verlag, Berlin-New York, 1972.

http://link.springer.com/book/10.1007%2FBFb0059108

last chapter

Exemples tirés de la géométrie algébrique

p 319 ex 1.2.6 a).

that states that the category of regular connections on $\mathbb P^1\backslash \{0,\infty\} $ admits $\mathbb G_a\times D_k(k/\mathbb Z)$ as Tannaka group, hence, if $k$ is algebraically closed of characteristic $0$, is non-canonically isomorphic to $\mathbb Z^{k, alg}$. This result also shows that the diagonalisable part is not compatible with base change. This explains why one is generally interested only by the unipotent fundamental group.