> does that mean that $x^2+x+1=0$ has the same roots as $0.\overline 9x^2+0.\overline 9x+0.\overline 9=0$
Yes, because that's two ways of representing exactly the same polynomial.
> If so, then does a given n-tuple of roots satisfy $2^{(n+1)}$ nth degree rational coefficient polynomials equated to zero?
This conclusion doesn't follow from the above, because the above just uses two ways of writing down the same polynomial.
I would actually argue that if an $n$-tuple is a root of an $n$th degree polynomial, then it is a root of infinitely many of them. Because if $p(x) = 0$ for some $x \in \Bbb R^n$, then $\alpha p(x) = 0$ as well, for that same $x$ and for all $\alpha \in \Bbb R$. Or take $\alpha \in \Bbb Q$ if you want to keep rational coefficients. Still infinitely many of them, and either way, the polynomial $\alpha p$ is different from $p$ as long as $\alpha \
e 1$.