What algebraic group does Tannaka-Krein reconstruct when fed the category of modules of a non-algebraic Lie algebra? Let $\mathfrak g$ be a finite-dimensional Lie algebra over $\mathbb C$, and let $\mathfrak g \text{-rep}$ be its category of finite-dimensional modules.  Then $\mathfrak g\text{-rep}$ comes equipped with a faithful exact functor "forget" to the category of finite-dimensional vector spaces over $\mathbb C$.  Moreover, $\mathfrak g\text{-rep}$ is symmetric monoidal with duals, and the forgetful functor preserves all this structure.  By Tannaka-Krein duality (see in particular the excellent paper André Joyal and Ross Street, An introduction to Tannaka duality and quantum groups, 1991), from this data we can reconstruct an affine algebraic group $\mathcal G$ such that $\mathfrak g \text{-rep}$ is equivalent (as a symmetric monoidal category with a faithful exact functor to vector spaces) to the category of finite-dimensional representations of $\mathcal G$.
However, it is not true that every finite-dimensional Lie algebra is the Lie algebra of an algebraic group.  So it is not true that $\mathcal G$ is, say, necessarily the simply-connected connected Lie group with Lie algebra $\mathfrak g$, or some quotient thereof.  So my question is:

Given $\mathfrak g$, what is an elementary description of $\mathcal G$ (that avoids the machinery of Tannaka-Krein)?

For example, perhaps $\mathcal G$ is some Zariski closure of something...?
 A: Rather than an answer, this is more of an anti-answer: I'll try to persuade you that you probably don't want to know the answer to your question.
Instead of some exotic nonalgebraic Lie algebra, let's start with the one-dimensional Lie algebra $\mathfrak{g}$ over a field $k$. A representation of $\mathfrak{g}$ is just a finite-dimensional vector space + an endomorphism. I don't know what the affine group scheme attached to this Tannakian category is, but thanks to a 1954 paper of Iwahori, I can tell you that its Lie algebra can be identified with the set of pairs $(\mathfrak{g},c)$ where $\mathfrak{g}$ is a homomorphism of abelian groups $k\to k$ and $c$ is an element of $k$. So if $k$ is big, this is huge; in particular, you don't get an algebraic group. (Added: $k$ is algebraically closed.)
By contrast, the affine group scheme attached to the category of representations of a semisimple Lie algebra $\mathfrak{g}$ in characteristic zero is the simply connected algebraic group with Lie algebra $\mathfrak{g}$. 
In summary: this game works beautifully for semisimple Lie algebras (in characteristic zero), but otherwise appears to be a big mess. See arXiv:0705.1348 for a few more details.  
A: 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.
