I saw this question a while ago and felt something in the way of a (probably misguided) missionary zeal to make at least a few elementary remarks. But upon reflection, it became clear that even that would end up rather long, so it was difficult to find the time until now.
The point to be made is a correction: fundamental groups in arithmetic geometry are not the same as Galois groups, per se. Of course there is a long tradition of parallels between Galois theory and the theory of covering spaces, as when Takagi writes of being misled by Hilbert in the formulation of class field theory essentially on account of the inspiration from Riemann surface theory. And then, Weil was fully aware that homology and class groups are somehow the same, while speculating that a sort of non-abelian number theory informed by the full theory of the 'Poincare group' would become an ingredient of many serious arithmetic investigations.
A key innovation of Grothendieck, however, was the formalism for refocusing attention on the base-point. In this framework, which I will review briefly below, when one says $$\pi_1(Spec(F), b)\simeq Gal(\bar{F}/F),$$ the base-point in the notation is the choice of separable closure $$b:Spec(\bar{F})\rightarrow Spec(F).$$ That is,
Galois groups are fundamental groups with generic base-points.
The meaning of this is clearer in the Galois-theoretic interpretation of the fundamental group of a smooth variety $X$. There as well, the choice of a separable closure $k(X)\hookrightarrow K$ of the function field $k(X)$ of $X$ can be viewed as a base-point $$b:Spec(K)\rightarrow X$$ of $X$, and then $$\pi_1(X,b)\simeq Gal(k(X)^{ur}/k(X)),$$ the Galois group of the maximal sub-extension $k(X)^{ur}$ of $K$ unramified over $X$. However, it would be quite limiting to take this last object as the definition of the fundamental group.
We might recall that even in the case of a path-connected pointed topological space $(M,b)$ with universal covering space $$M'\rightarrow M,$$ the isomorphism $$Aut(M'/M)\simeq \pi_1(M,b)$$ is not canonical. It comes rather from the choice of a base-point lift $b'\in M'_b$. Both $\pi_1(M,b)$ and $Aut(M'/M)$ act on the fiber $M'_b$, determining bijections $$\pi_1(M,b)\simeq M'_b\simeq Aut(M'/M)$$ via evaluation at $b'$. It is amusing to check that the isomorphism of groups obtained thereby is independent of $b'$ if and only if the fundamental group is abelian. The situation here is an instance of the choice involved in the isomorphism $$\pi_1(M,b_1)\simeq \pi_1(M,b_2)$$ for different base-points $b_1 $ and $b_2$. The practical consequence is that when fundamental groups are equipped with natural extra structures coming from geometry, say Hodge structures or Galois actions, different base-points give rise to enriched groups that are are often genuinely non-isomorphic.
A more abstract third group is rather important in the general discussion of base-points. This is $$Aut(F_b),$$ the automorphism group of the functor $$F_b:Cov(M)\rightarrow Sets$$ that takes a covering $$N\rightarrow M$$ to its fiber $N_b$. So elements of $Aut(F_b)$ are compatible collections $$(f_N)_N$$ indexed by coverings $N$ with each $f_N$ an automorphism of the set $N_b$. Obviously, newcomers might wonder where to get such compatible collections, but lifting loops to paths defines a natural map $$\pi_1(M,b)\rightarrow Aut(F_b)$$ that turns out to be an isomorphism. To see this, one uses again the fiber $M'_b$ of the universal covering space, on which both groups act compatibly. The key point is that while $M'$ is not actually universal in the category-theoretical sense, $(M',b')$ is universal among pointed covers. This is enough to show that an element of $Aut(F_b)$ is completely determined by its action on $b'\in M'_b$, leading to another bijection $$Aut(F_b)\simeq M'_b.$$ Note that the map $\pi_1(M,b)\rightarrow Aut(F_b)$ is entirely canonical, even though we have used the fiber $M'_b$ again to prove bijectivity, whereas the identification with $Aut(M'/M)$ requires the use of $(M'_b,b')$ just for the definition.
Among these several isomorphic groups, it is $Aut(F_b)$ that ends up most relevant for the definition of the etale fundamental group.
So for any base-point $b:Spec(K)\rightarrow X$ of a connected scheme $X$ (where $K$ is a separably closed field, a 'point' in the etale theory), Grothendieck defines the 'homotopy classes of etale loops' as $$\pi^{et}_1(X,b):=Aut(F_b),$$ where $$F_b:Cov(X) \rightarrow \mbox{Finite Sets}$$ is the functor that sends a finite etale covering $$Y\rightarrow X$$ to the fiber $Y_b$. Compared to a construction like $Gal(k(X)^{ur}/k(X))$, there are three significant advantages to this definition.
(1) One can easily consider small base-points, such as might come from a rational point on a variety over $\mathbb{Q}$.
(2) It becomes natural to study the variation of $\pi^{et}_1(X,b)$ with $b$.
(3) There is an obvious extension to path spaces $$\pi^{et}_1(X;b,c):=Isom(F_b,F_c),$$ making up a two-variable variation.
This last, in particular, has no analogue at all in the Galois group approach to fundamental groups. When $X$ is a variety over $\mathbb{Q}$, it becomes possible, for example, to study $\pi^{et}_1(X,b)$ and $\pi^{et}_1(X;b,c)$ as sheaves on $Spec(\mathbb{Q})$, which encode rich information about rational points. This is a long story, which would be rather tiresome to expound upon here (cf. lecture at the INI ). However, even a brief contemplation of it might help you to appreciate the arithmetic perspective that general $\pi^{et}_1$'s are substantially more powerful than Galois groups. Having read thus far, it shouldn't surprise you that I don't quite agree with the idea explained, for example, in this post that a Galois group is only a 'group up to conjugacy'. To repeat yet again, the usual Galois groups are just fundamental groups with specific large base-points. The dependence on these base-points as well as a generalization to small base-points is of critical interest.
Even though the base-point is very prominent in Grothendieck's definition, a curious fact is that it took quite a long time for even the experts to fully metabolize its significance. One saw people focusing mostly on base-point independent constructions such as traces or characteristic polynomials associated to representations. My impression is that the initiative for allowing the base-points a truly active role came from Hodge-theorists like Hain, which then was taken up by arithmeticians like Ihara and Deligne. Nowadays, it's possible to give entire lectures just about base-points, as Deligne has actually done on several occasions.
Here is a puzzle that I gave to my students a while ago: It has been pointed out that $Gal(\bar{F}/F)$ already refers to a base-point in the Grothendieck definition. That is, the choice of $F\hookrightarrow \bar{F}$ gives at once a universal covering space and a base-point. Now, when we turn to the manifold situation $M'\rightarrow M$, a careful reader may have noticed a hint above that there is a base-point implicit in $Aut(M'/M)$ as well. That is, we would like to write $$Aut(M'/M)\simeq \pi_1(M,B)$$ canonically for some base-point $B$. What is $B$?
Added:
-In addition to the contribution of Hodge-theorists, I should say that Grothendieck himself urges attention to many base-points in his writings from the 80's, like 'Esquisse d'un programme.'
-I also wanted to remark that I don't really disagree with the point of view in JSE's answer either.
Added again:
This question reminds me to add another very basic reason to avoid the Galois group as a definition of $\pi_1$. It's rather tricky to work out the functoriality that way, again because the base-point is de-emphasized. In the $Aut(F_b)$ approach, functoriality is essentially trivial.
Added, 27 May:
I realized I should fix one possible source of confusion. If you work it out, you find that the bijection $$\pi_1(M,b)\simeq M'_b\simeq Aut(M'/M)$$ described above is actually an anti-isomorphism. That is, the order of composition is reversed. Consequently, in the puzzle at the end, the canonical bijection $$Aut(M'/M)\simeq \pi_1(M,B)$$ is an anti-isomorphism as well. However, another simple but amusing exercise is to note that the various bijections with Galois groups, like $$\pi_1(Spec(F), b)\simeq Gal(\bar{F}/F),$$ are actually isomorphisms.
Added, 5 October:
I was asked by a student to give away the answer to the puzzle. The crux of the matter is that any continuous map $$B:S\rightarrow M$$ from a simply connected set $S$ can be used as a base-point for the fundamental group. One way to do this to use $B$ to get a fiber functor $F_B$ that associates to a covering $$N\rightarrow M$$the set of splittings of the covering $$N_B:=S\times_M N\rightarrow S$$ of $S$. If we choose a point $b'\in S$, any splitting is determined by its value at $b'$, giving a bijection of functors $F_B=F_{b'}=F_b$ where $b=B(b')\in M$. Now, when $$B:M'\rightarrow M$$ is the universal covering space, I will really leave it as a (tautological) exercise to exhibit a canonical anti-isomorphism $$Aut(F_B)\simeq Aut(M'/M).$$ The 'point' is that $$F_B(M')$$ has a canonical base-point that can be used for this bijection.
