You could do it sort of without much field theory. The only result I'm going to use is a baby version of the fundamental theorem of symmetric function:
If $P$ is any symmetric function in $n$ variables and $\alpha_1,\dots,\alpha_n$ are roots of a monic polynomial $x^n+a_{n-1}x^{n-1}+\dots+a_0$ then $P(\alpha_1,\dots,\alpha_n)=Q(a_{n-1},a_{n-2},\dots,a_0)$ for some polynomial $Q$ whose coefficients are $\mathbb{Z}$-linear combinations of the coefficients of $P$.
The proof of this version is basically the same as other versions: you show that every monomial symmetric function can be written this way by induction on (reverse) lexicographic ordering of degrees. We drop uniqueness and the ring language in this version because they are not needed.
Suppose $\alpha$ is a root of $f(x)=x^n+\alpha_{n-1}x^{n-1}+\dots+\alpha_0$ where $\alpha_i$ are algebraic. That means we have monic polynomials $g_{n-1},\dots,g_0$ with rational coefficients such that $g_k(\alpha_k)=0$. Let $g_k$ has roots $\beta_{k,j}$, $j=1,2,\dots,\deg g_k$ (for definiteness, say $\beta_{k,1}=\alpha_k$) and we look at the polynomial in $n+1$ variables
$$
F(X_n,X_{n-1},\dots,X_0)=\prod_{\substack{1\leq i_0\leq\deg g_0\\\vdots\\1\leq i_{n-1}\leq\deg g_{n-1}}} \Big(X_n+\beta_{n-1,i_{n-1}}X_{n-1}+\dots+\beta_{0,i_{n_0}}X_0\Big)
$$
We claim this has rational coefficients when expanded. Indeed, each coefficient is a symmetric polynomial in the $\beta_{i,j}$ for all $i$, so inductively replace dependence on $(\beta_{i,j})_{j=1}^{\deg g_i}$ for each $i$, we end up with every coefficient being an integer-coefficient polynomial in the coefficients of $g_i$. So $F$ has rational coefficients. But now
$$
g(X)=F(X^n,X^{n-1},\dots,1)=\prod_{\substack{1\leq i_0\leq\deg g_0\\\vdots\\1\leq i_{n-1}\leq\deg g_{n-1}}} \Big(X^n+\beta_{n-1,i_{n-1}}X^{n-1}+\dots+\beta_{0,i_{n_0}}\Big)
$$
is a monic polynomial with rational coefficients and has a factor $f(X)$, hence has $\alpha$ as a root. QED.
I don't think you could get away with using anything more "elementary" than this.
Remark: You could jump directly to $g(X)$ without going through $F$ but I think this way makes the dependence on $\beta$s easier to track -- the coefficient of $X_n^{m_n}\dots X_0^{m_0}$ is the sum of the product of $m_i$ of $\beta_{i,j}$ over all possible choices of $j$s with the multiplicity bound: each $j$ can only appear $\leq \frac{\prod_{i'=1}^{n-1}\deg g_{i'}}{\deg g_i}$ times.
