This is a great question. Someone will come along with a better answer I'm sure, but here's a bit off the top of my head:
1) The Hilbert class field of a number field $K$ is the maximal everywhere unramified abelian extension of $K$. (Here when we say "$K$" we really mean "$\mathbb{Z}_K$", the ring of integers. That's important in the language of etale maps, because any finite separable field extension is etale.)
In the case of a curve over $\mathbb{C}$, the "problem" is that there are infinitely many unramified abelian extensions. Indeed, Galois group of such is the abelianization of the fundamental group, which is free abelian of rank $2g$ ($g$ = genus of the curve). Let me call this group G.
This implies that the covering space of C corresponding to G has infinite degree, so is a non-algebraic Riemann surface. In fact, I have never really thought about what it looks like. It's fundamental group is the commutator subgroup of the fundamental group of C, which I believe is a free group of infinite rank. I don't think the field of meromorphic functions on this guy is what you want.
2) On the other hand, the Hilbert class group $G$ of $K$ can be viewed as the Picard group of $\mathbb{Z}_K$, which classifies line bundles on $\mathbb{Z}_K$. This generalizes nicely: the Picard group of $C$ is an exension of $\mathbb{Z}$ by a $g$-dimensional complex torus $J(C)$, which has exactly the same abelian fundamental group as $C$ does: indeed their first homology groups are canonically isomorphic. $J(C)$ is called the Jacobian of $C$.
3) It is known that every finite unramified abelian covering of $C$ arises by pulling back an isogeny from $J(C)$.
So there are reasonable claims for calling either $G \cong \mathbb{Z}^{2g}$ and $J(C)$ the Hilbert class group of $C$. These two groups are -- canonically, though I didn't explain why -- Pontrjagin dual to each other, whereas a finite abelian group is (non-canonically) self-Pontrjagin dual. [This suggests I may have done something slightly wrong above.]
As to what the Hilbert class field should be, the analogy doesn't seem so precise. Proceeding most literally you might take the direct limit of the function fields of all of the unramified abelian extensions of $C$, but that doesn't look like such a nice field.
Finally, let me note that things work out much more closely if you replace $\mathbb{C}$ with a finite field $\mathbb{F}_q$. Then the Hilbert class field of the function field of that curve is a finite abelian extension field whose Galois group is isomorphic to $J(C)(\mathbb{F}_q)$, the (finite!) group of $\mathbb{F}_q$-rational points on the Jacobian.
As in Birdman's comment, the motivic fundamental group is unifying the notion of monodromy action on the fibers of local systems of "geometric origin."
To explain this, let us start with the case of a field $K$. We have a semisimple $\mathbb{Q}$-linear Tannakian category $\operatorname{Mot}_K$ of (pure) motives over $K$ for which fiber functors are cohomology theories, i.e., it makes sense to have an $L$-valued fiber functor for a field $L$, and this is the same as a Weil cohomology theory for smooth proper $K$-varieties with values in $L$-vector spaces. A motivic Galois group, to my understanding, is attached to a cohomology theory/fiber functor $F$ of $\operatorname{Mot}_K$.
Then the motivic Galois group is the associated group scheme/$L$ whose representations are given by the category $\operatorname{Mot}_{K}\underset{\mathbb{Q}}{\otimes}L$, i.e., it is the group scheme of automorphisms of the fiber functor $F$. So it is "the group which acts on $F$-cohomology of (smooth projective) varieties." Since this category is semi-simple, the motivic Galois group is pro-reductive. E.g., the absolute Galois group (considered as a discrete group scheme) of $K$ acts on $\ell$-adic cohomology, so there is a homomorphism from $\operatorname{Gal}(K)$ to the motivic Galois group of $K$ corresponding to the fiber functor defined by $\ell$-adic cohomology.
For, say, a smooth variety $X$ over $K$, there should be a category of "motivic sheaves" on $X$, or rather, a semi-simple category of pure motivic sheaves contained in an Artinian category of mixed motivic sheaves. You should have e.g. an $\ell$-adic" fiber functor from the mixed category to $\ell$-adic perverse sheaves on $X$ which sends pure guys to (cohomologically shifted) lisse sheaves (alias local systems). E.g., if $K=\mathbb{F}_q$, then this is the category of pure (resp. mixed) perverse sheaves on $X$. If $K=\mathbb{C}$, this should be a full subcategory of pure (resp. mixed) polarizable Hodge modules on $X$. For any smooth proper (resp. just any) map $f:Y\to X$, there should an object in the category of pure (resp. mixed) motivic sheaves on $X$ corresponding to push-forward of the structure sheaf on $Y$.
The motivic fundamental group act on the ``fibers" of pure motivic sheaves on $X$. I.e., for a $K$-point of $X$, you should get a functor to the category of $K$-motives. This is a motivic incarnation of taking the fiber of a local system. Then given our cohomology theory $F$, we obtain a functor from pure motivic sheaves on $X$ to $L$-vector spaces, and the automorphisms of this functor will be the $F$-realization of the motivic Galois group of $X$.
Best Answer
Essentially by definition, the etale fundamental group of a field $k$ is the group of $k$-automorphisms of a separable algebraic closure $k^{\operatorname{sep}}$ over $k$. If $k$ is perfect, then this is (well-defined up to an inner automorphism) $\operatorname{Aut}(\overline{k}/k)$. On the other hand, any $k$-automorphism of $k^{\operatorname{sep}}$ extends uniquely to an automophism of the algebraic closure $\overline{k}$. Therefore in every case the etale fundamental group of $k$ is isomoprhic to $\operatorname{Aut}(\overline{k}/k)$.