Yes, there even exist symmetric polynomials with such properties.
Let $\sqrt{2}=c_1<c_2<c_3<\ldots$ be a bounded sequence. Our aim is to construct inductively symmetric polynomials $f_1,\ldots,f_n$, on the spheres $S^1,\ldots,S^n$ respectively such that $f_1(a_1,a_2)=a_1+a_2$, $\max |f_i|\leqslant c_i$ and $f_n(a_1,\ldots,a_{n+1})=f_{n+1}(a_1,\ldots,a_{n+1},0)$.
Assume that $f_n$ is constructed. Denote $\Delta=\{(x_1,\ldots,x_{n+2})\in S^{n+2}:\prod x_i=0\}$.
First of all, construct some symmetric polynomial $G(x_1,\ldots,x_{n+2})$ such that $G(x_1,\ldots,x_{n+1},0)=f_n(x_1,\ldots,x_{n+1})$ (this is easy to do by symmetrizing all monomials of $f_n$.)
Next, construct a symmetric continuous function $h(x_1,\ldots,h_{n+2})$ which coincides with $G$ on a small neighborhood $V$ of the set $\Delta$ and satisfies $\max |h|\leqslant (c_n+c_{n+1})/2$. There are many ways to do it, for example by solving Dirichlet Laplacian problem on $S^{n+1}\setminus V$, we only need to choose $V$ so that $|G|<(c_n+c_{n+1})/2$ on $V$.
Finally denote $p=\frac{G-h}{x_1\ldots x_{n+2}}$. This is a continuous symmetric function on $S^{n+2}$ (vanishing on $V$), it may be approximated by a symmetric polynomial $q$ uniformly on $S^{n+2}$ with accuracy $(c_{n+1}-c_n)/2$. Put $f_{n+1}=G-q\cdot \prod x_i$. It is a symmetric polynomial coinciding with $G$ on $\Delta$ and satisfying
$$
|f_{n+1}(x_1,\ldots,x_{n+2})|=|G-q\prod x_i|\leqslant |G-p\prod x_i|+|p-q|\leqslant \max |h|+\max |p-q|\leqslant c_{n+1}.
$$