I don’t know how to make use of the full power of the intermediate value theorem, hence I will work instead with a weaker axiomatization: let $\def\rcf{\mathrm{RCF}}\rcf'_k$ denote the first two groups of axioms from the question (i.e., $\rcf_1$) + the assertion that every polynomial of odd degree $d\le k$ has a root, and let
$$\def\fF{\mathfrak F}\fF'(n)=\min\{k:\rcf'_k\vdash\theta_n\},\quad n>2.$$
Since $\rcf_k\vdash\rcf'_k$, we have $\fF(n)\le\fF'(n)$. (On the other hand, $\bigwedge_{n=3}^k\theta_n$ implies $\rcf_k$ over $\rcf_1$, hence $\rcf'_{\max\{\fF'(n):3\le n\le k\}}\vdash\rcf_k$. I don’t know if one can do better.)
I claim that if $n$ is not a power of $2$, then
$$\fF(n)\le\fF'(n)\le\binom n{n_2},\quad\text{where }n_2=2^{v_2(n)}$$
(i.e., $n_2$ is the highest power of $2$ that divides $n$). Note that $\binom n{n_2}$ is odd (see e.g. Kummer’s theorem).
To see this, let $R\models\rcf'_k$ for $k=\binom n{n_2}$, and assume for contradiction that $f\in R[x]$ of degree $n$ is irreducible. Let $\{\alpha_i:i\in[n]\}$ list the roots of $f$ in a fixed splitting field of $f$ over $R$.
For any symmetric polynomial $s\in R[x_1,\dots,x_{n_2}]$, the coefficients of
$$p_s(x)=\prod_{i_1<\dots<i_{n_2}}(x-s(\alpha_{i_1},\dots,\alpha_{i_{n_2}}))$$
are symmetric polynomials in the $\alpha_i$’s, hence $p_s\in R[x]$. Since $p_s$ has degree $k$, it has a root in $R$: i.e., there is $\{i_1,\dots,i_{n_2}\}\subseteq[n]$ such that
$$s(\alpha_{i_1},\dots,\alpha_{i_{n_2}})\in R.$$
Let $\{\sigma_t:t\le n_2\}$ denote the elementary symmetric polynomials in $n_2$ variables. For each $I=\{i_1,\dots,i_{n_2}\}\subseteq[n]$, $i_1<\dots<i_{n_2}$, the set
$$V_I=\bigl\{(r_1,\dots,r_{n_2})\in R^{n_2}:\textstyle\sum_tr_t\sigma_t(\alpha_{i_1},\dots,\alpha_{i_{n_2}})\in R\bigr\}$$
is an $R$-linear space, and by what we have just proved above,
$$R^{n_2}=\bigcup_{\substack{I\subseteq[n]\\|I|=n_2}}V_I.$$
Since a linear space over an infinite field is not a finite union of proper subspaces (see e.g. this question), there is $I=\{i_1,\dots,i_{n_2}\}$ such that $V_I=R^{n_2}$. That is, $\sigma_t(\alpha_{i_1},\dots,\alpha_{i_{n_2}})\in R$ for all $t=1,\dots,n$, which means that
$$\prod_{j=1}^{n_2}(x-\alpha_{i_j})\in R[x],$$
contradicting the irreducibility of $f$ (as $n_2<n$ by assumption).
Note that what we have really shown is that $\rcf'_k$ proves that every polynomial of degree $n$ has a factor of degree $n_2$.
What about the case when $n$ is a power of $2$? In the comments, I gave an argument that
$$\fF'(n)\le\frac{n!}{2^{v_2(n!)}}=\frac{n!}{2^{n-1}},$$
but this is considerably larger than the bound in the non-power-of-$2$ case. Let me outline how to prove an asymptotically better bound
$$\fF'(n)\le\binom{\binom n{n/2}}2.$$
Let $\def\acf{\mathrm{ACF}}\acf_k$ denote the theory of fields such that every polynomial of degree $2$ or of odd degree $d\le k$ has a root.
If $R\models\rcf'_{\binom{2k}2}$ then $R(i)\models\acf_k$: $\rcf'_1$ already implies that $R(i)$ is quadratically closed, and if $f\in R(i)[x]$ is a degree-$d$ polynomial with $d$ odd, then $f\overline f\in R[x]$ has degree $2d$, hence as we have shown, $\rcf'_{\binom{2d}2}$ ensures that it has a quadratic factor in $R[x]$, hence a root in $R(i)$.
By the same reasoning as above, $\acf_{\binom n{n_2}}$ proves that any degree-$n$ polynomial has a degree-$n_2$ factor. In particular, if $n\equiv2\pmod4$, then $\acf_{\binom n2}$ proves that any degree-$n$ polynomial has a root.
If $n$ is a power of $2$, then $\acf_{\binom{\binom n{n/2}}2}$ proves that any polynomial $f$ of degree $n$ has a factor of degree $n/2$, hence, by induction on $n$, it proves that any such polynomial has a root: we do the construction above with $n/2$ in place of $n_2$; since $\binom n{n/2}\equiv2\pmod4$, all the degree-$\binom n{n/2}$ polynomials considered have roots by the previous paragraph.
Now, let $n\ge4$ be a power of $2$, and $f\in R[x]$ of degree $n$, where $R\models\rcf'_{\binom{\binom n{n/2}}2}$. Doing again the symmetric polynomial construction with $n/2$ in place of $n_2$, each of the relevant degree-$\binom n{n/2}$ polynomials has a quadratic factor over $R$, hence a root in $R(i)$. In this way, we find a degree-$n/2$ polynomial $g\in R(i)[x]$ such that $g\mid f$. Since $2\binom{\binom{n/2}{n/4}}2\le\binom n{n/2}$, we have $R(i)\models\acf_{\binom{\binom{n/2}{n/4}}2}$, thus by the previous paragraph, $g$ has a root in $R(i)$, hence $f$ has a root or a quadratic factor over $R$.
