Here is the proof of the equivalent statement "Every complex non-constant polynomial $p$ is surjective".
1) Let $C$ be the finite set of critical points , i.e. $p'(z)=0$ for all $z\in C$. $C$ is finite by elementary algebra.
2) Remove $p(C)$ from the codomain and call the resulting open set $B$ and remove from the domain its inverse image $p^{-1}\left( p (C) \right)$, and call the resulting open set $A$. Note that the inverse image is again finite.
3) Now you get an open map from $A$ to $B$, which is also closed, because any polynomial is proper (inverse images of compact sets are compact). But $B$ is connected and so $p$ is surjective.
I like this proof because you can try it for real polynomials and it breaks down at step 3) because if you remove a single point from the line you disconnect it, while you can remove a finite set from a plane leaving it connected.
I think I can now reasonably point to the book
Bourbaki, Topologie Algébrique: Chapitres 1 à 4 (Elements de Mathématique) 7 Apr 2016
which does use ideas of descent and the fundamental groupoid both for the van Kampen theorem and for discrete groups acting properly on spaces. They call the fundamental groupoid the "Poincaré groupoid", though I think its topological use goes back only to Reidemeister's 1932 book. However, despite the comment of Grothendieck, they do not use the fundamental groupoid on a set of base points.
Their Theorem 3 on p. 419 seems to overlap with the work on orbit groupoids of Chapter 11 of Topology and Groupoids, which gives the application to the symmetric square of a space.
Comments welcome!
Aug 21, 2016 I should mention the paper
"Van Kampen theorems for categories of covering
morphisms in lextensive categories" R. Brown, G. Janelidze, J. Pure Applied Algebra I 19 (1997) 255-283. (pdf)
This considers the whole fundamental groupoid, not "many base points", but does use descent notions in general situations.
Best Answer
As far as I'm aware, there isn't a compelling direct argument from Brouwer fixed point to imply the fundamental theorem of algebra. Such an argument isn't impossible -- I can imagine some fairly contrived proofs but I don't know of a very natural one. The references like Guillemin and Pollack don't derive FTOA from Brouwer, they derive both FTOA and Brouwer from degree/intersection theory. In particular they only use mod-2 degree theory for Brouwer but oriented degree theory for FTOA.
I had an argument written down here previously that I thought might work but now I realize it can't work. Oh, but it's fixable.
EDIT I've managed to repair the argument. The downside is it's not as simple.
A polynomial without roots produces a polynomial without fixed points. Specifically, $p(z) \neq 0$ for all $z \in \mathbb C$ means $q(z) = p(z)+z$ has no fixed points in $\mathbb C$. So what? Think of $q(z)$ as a map of the Riemann sphere. Now take the real oriented blow-up of the Riemann sphere at infinity (i.e. replace the point at infinity by its unit normal bundle in the sense of smooth real manifolds). This is a disc. So $q(z)$ becomes a smooth map of the disc, denote it $\hat q$, and identify the blow-up with $D^2$, the unit disc in $\mathbb C$ centred at the origin.
If $q(z) = z^n + a_{n-1}z^{n-1} + \cdots + a_1 z + a_0$ and if we conjugate by $z \longmapsto 1/z$ so
$$\frac{1}{q(1/z)} = \frac{z^n}{1+a_{n-1}z + \cdots + a_0z^n}$$
So when you restrict $\hat q$ to the boundary circle, it becomes $z \longmapsto z^n$.
$z \longmapsto z^n$ has fixed points $z^{n-1}=1$, the $(n-1)$-th roots of unity. So we can not directly appeal to Brouwer, since Brouwer's fixed point theorem might give you a pre-existing fixed point on the boundary.
Consider the vector field $v(z) = z-\hat q(z)$ on $D^2$. It is inward-pointing on the boundary circle with the sole exception of $z^{n-1}=1$, the $(n-1)$ roots of unity. But if we remove a small neighbourhood of $\partial D^2$ from $D^2$, the vector field $v$ restricts to an inward pointing vector field. So you could appeal to Poincare-Hopf and say there has to be a zero in the interior, or you could talk about the flow of the vector field, and Brouwer's fixed-point theorem would then tell you the vector field must have a zero in the interior.
So its not a slick proof, but it can be done.
A suitable identification between $D^2$ and the blow-up of the Riemann sphere at infinity is done by the map $X : D^2 \to \hat{\mathbb C}$ given by $X(z) = \frac{1}{1-|z|^2} z$. So $\hat q$ is the unique continuous extension of $X \circ q \circ X^{-1}$ to $D^2$.