Is there a purely algebraic proof of the Fundamental Theorem of Algebra?

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

Such a proof was given in the following published article in a refereed journal:

P. Blaszczyk, A Purely Algebraic Proof of the Fundamental Theorem of Algebra. Annales Universitatis Paedagogicae Cracoviensis Studia ad Didacticam Mathematicae Pertinentia VIII, 2016, 5-21.

Thus the name of the theorem (FTA) is fully justified. The proof uses an ultrapower.