After some thought my pessimism (as expressed in my concurrence with the answer
of Milne) has abated somewhat. If I were bold enough I would conjecture the
following (assuming that the characteristic zero base field is algebraically
closed): Let $\mathfrak g$ be a finite dimensional Lie algebra over $k$ and let
$G$ be the pro-algebraic group whose representation tensor category is
equivalent to the tensor category of finite dimensional $\mathfrak
g$-modules. Then if $S$ is the (pro-)radical of $G$ and $U$ the (pro-)unipotent
radical $U$ and $G/S$ are algebraic groups (unipotent and semi-simple
respectively). Furthmore, the pro-torus $T:=S/U$ has as character group
$\mathfrak{u}/[\mathfrak{g},\mathfrak{u}]$ considered as an additive
group. Hence the only infinite-dimensional part is $T$ but its character group,
\emph{à priori} only an abstract group, is reasonably well controlled. This is
analogous to the case of irreducible infinite-dimensional
representations of a semi-simple Lie group where the center of the enveloping
algebra acts by a character and the set of characters as a set is very
large. However it is the set of $k$-points of an algebraic variety which means
that it is under control. The analogy goes further as the category of
$G$-representations (assuming $U$ is finite dimensional) splits up into a direct
product of categories parametrised by cosets of the character group of $T$ with
respect to the subgroup generated by the characters occurring in the action of
$T$ on the Lie algebra of $U$.
Here are some comments on the conjecture (I do not vouch for the complete
veracity of my claims).
We can get a picture of $G/U$ by looking at the irreducible $\mathfrak
g$-representations (as they correspond exactly to the irreducible
$G/U$-representations). All such representations factor through $\mathfrak
g/[\mathfrak{g},\mathfrak{u}]$ which is the product of
$\mathfrak{u}/[\mathfrak{g},\mathfrak{u}]$ and $\mathfrak g/\mathfrak{u}$.
Hence, the irreducible representations are parametrised by pairs of a
$1$-dimensional representation of $\mathfrak{u}/[\mathfrak{g},\mathfrak{u}]$ and
an irreducible representation of the semi-simple algebra $\mathfrak
g/\mathfrak{u}$. This gives the prediction that $G/U$ should be the product of a
torus with character group $\mathfrak{u}/[\mathfrak{g},\mathfrak{u}]$ and the
simply-connected semi-simple group with Lie algebra $\mathfrak g/\mathfrak{u}$.
As for $U$, the idea is that the category of
unipotent representations (i.e., successive extensions of the trivial
representation) of $\mathfrak g$ is equivalent to the category of
representations of the unipotent group with Lie algebra $\mathfrak g'$, the
maximal unipotent quotient of $\mathfrak g$. Something similar ought to be true
for successive extension of the same irreducible representation and there
shouldn't be too much "intermixing" between different irreducibles.
[Added]
I somewhat rudely hijacked the question by taking up things that maybe weren't that
pertinent to the question so let me give an answer which I think is more on track.
The problem is that one can not always define the algebraisation of an
abstract finite dimensional Lie algebra $\mathfrak g$ even if some
algebraisation exists. As an examples consider a $2$-dimensional Lie algebra
with basis $x,y$ and $[x,y]=y$. This is the Lie algebra of an infinite number of
algebraic groups: Let the $1$-dimensional torus $\mathbb G_m$ act on the
additive group $\mathbb G_a$ by $(t,v) \mapsto t^nv$, where $n\not=0$ and let $G_n$ be the
semi-direct product of this action. These groups all have $\mathfrak g$ as Lie
algebra but the only isomorphisms between them is that $G_n$ is isomorphic to
$G_{-n}$.
What does make sense is to speak of an algebraic hull of an embedding of
$\mathfrak g\subseteq \mathfrak{gl}_m$, i.e., of a (faithful) $\mathfrak
g$-representation. In that case one may consider the intersection of all
algebraic subgroups of $\mathrm{GL}_m$ whose Lie algebra contains $\mathfrak
g$. In terms of Zariski closures (when the base field is $\mathbb C$) it is the Zariski closure of the
exponentials of all elements of $\mathfrak g$ (inside of
$\mathfrak{gl}_m$). From the Tannakian point of view this is the group that
corresponds to the tensor subcategory of the category of $\mathfrak
g$-representations generated by the given representation.
However, if one wants something that is independent of a particular
representation one has to pass to an inverse limit of groups coming from
different representations. This leads to an infinite dimensional monster even in
the case when $\mathfrak g$ is $1$-dimensional.
