No, there is no purely algebraic proof of FTA.
So, as someone already noted, FTA is a misnomer.
I think the following proof is one of the most algebraic ones, though it's not purely algebraic.
Assumptions
We assume the following facts.
(1) Every polynomial of odd degree in $\mathbb{R}[X]$ has a root in $\mathbb{R}$.
(2) Every polynomial of degree 2 in $\mathbb{C}[X]$ has a root in $\mathbb{C}$.
Note:
(1) can be proved by the intermediate value theorem.
(2) can be proved by the fact that every polynomial of degree 2 in $\mathbb{R}[X]$ has a root in $\mathbb{C}$.
Notation
We denote by $|G|$ the order of a finite group $G$.
Lemma
Let $K$ be a field.
Suppose every polynomial of odd degree in $K[X]$ has a root in K.
Let $L/K$ be a finite Galois extension.
Then the Galois group $G$ of $L/K$ is a 2-group.
Proof:
We can assume that $L \neq K$.
By the theorem of primitive element, there exists $\alpha$ such that $L = K(\alpha)$.
By the assumption, the degree of the minimal polynomial of $\alpha$ is even.
Hence $|G|$ is even.
Let $|G| = 2^r m$, where $m$ is odd.
Let $P$ be a Sylow 2-subgroup of $G$.
Let $M$ be the fixed subfield by $P$.
Since $(M : K) = m$ is odd, $m = 1$ by the similar reason as above.
QED
The fundamental theorem of algebra
The field of complex numbers $\mathbb{C}$ is algebraically closed.
Proof:
Let $f(X)$ in $\mathbb{R}[X]$ be non-constant.
It suffices to prove that $f(X)$ splits in $\mathbb{C}$.
Let $L/\mathbb{C}$ be a splitting field of $f(X)$.
Since $L/\mathbb{R}$ is a splitting field of $(X^2 + 1)f(X)$, $L/\mathbb{R}$ is Galois.
Let $G$ be the Galois group of $L/\mathbb{R}$.
Let $H$ be the Galois group of $L/\mathbb{C}$.
By the assumption (1) and the lemma, $G$ is a 2-group.
Hence $H$ is also a 2-group.
Suppose $|H| > 1$.
Since $H$ is solvable, $H$ has a nomal subgroup $N$ such that $(H : N) = 2$.
Let $F$ be the fixed subfield by $N$.
Since $(F : C) = 2$, this is a contradiction by the assumption (2).
Hence $H = 1$. It means $L = \mathbb{C}$.
QED
this is borrowed from the comments in the link to the question you posted:
Let $F$ be a field of characteristic $0$, and $K / F$ be a finite Galois extension. Suppose every polynomial of odd degree in $F[x]$ has a root in $F$, and every polynomial of degree $2$ in $K[x]$ has a root in $K$. Then $K$ is algebraically closed.
Also, it turns out that algebraically closures are rarely finite extensions. In fact, if $F$ is a field and $C / F$ a finite extension that is algebraically closed, then $C=F(i)$ where $i^{2}=-1$, and $F$ has characteristic $0$. Also for every nonzero $a \in F$, either $a$ or $-a$ is a square, and finite sums of nonzero squares are nonzero squares. The source for this is Keith Conrad's notes:
http://www.math.uconn.edu/~kconrad/blurbs/galoistheory/artinschreier.pdf
Best Answer
I can immediately think of four important areas for which the FTA is actually quite fundamental. Maybe other people may come up with more contributions.
Algebraic Geometry (as already touched in Agusti Roig's answer). In particular, we should couple the FTA with Lefschetz's Principle which basically says that the complex projective space is sort of "universal" environment for algebraic geometry in characteristic 0.
Classification of algebraic structures over local or global fields, where one uses constantly that $\Bbb C$ has no non trivial finite extensions (a trivial consequence of the FTA)
The theory of representations of groups, where the TFA plays a big role making the theory of complex representations simpler.
Galois theory and algebraic number theory, where the FTA allows to realize $\Bbb C$ as a good environment for studying finite extensions of $\Bbb Q$ (such in the basic resut that a number field of degree $n$ always admits $n$ independent embedings into $\Bbb C$) and allowing the use of the "geometry of numbers" (Minkowski's theorem) to prove important arithmetic results such as the finiteness of the class number and the structure of the units (Dirichlet's theorem)