I think that I can answer the first one of the three questions (which I guess is the simplest one), about the flat dimension. The idea is to reduce the problem to the case when $R$ is a field, using the classical Hilbert's syzygy theorem as a black box. The argument proceeds by Noetherian induction in the ring $R$. So we assume that the assertion is true for all the quotient rings of $R$ by nonzero ideals, and prove that it is also true for $R$.
Consider two cases separately.
I. $R$ has zero-divisors. So there are two nonzero elements $a$ and $b$ in $R$ such that $ab=0$.
We have an $R$-flat $S$-module $G$. Then $G/aG$ is an $R/aR$-flat $S/aS$-module and $G/bG$ is an $R/bR$-flat $S/bS$-module. By the assumption of Noetherian induction applied to the rings $R/aR$ and $R/bR$, the flat dimensions of the $S/aS$-module $G/aG$ and the $S/bS$-module $G/bG$ do not exceed $n$.
We have to show that $\operatorname{Tor}^S_{n+1}(M,G)=0$ for all $S$-modules $M$. Using the short exact sequence $0\to aM \to M \to M/aM \to 0$, the problem is reduced to the particular cases when either $aM=0$ or $bM=0$. Assume $aM=0$. Tensoring a flat resolution of the $S$-module $G$ with $R/aR$ over $R$, we obtain a flat resolution of the $S/aS$-module $G/aG$. Hence $\operatorname{Tor}^S_i(M,G)=\operatorname{Tor}^{S/aS}_i(M,G/aG)=0$ for all $i>n$, as desired.
II. $R$ is an integral domain. Denote by $Q$ the field of fractions of $R$.
Let $M$ be an $S$-module. We have $\operatorname{Tor}^S_i(M,G)\otimes_RQ\simeq\operatorname{Tor}^{S\otimes_RQ}_i(M\otimes_RQ,\>G\otimes_RQ)$ for all $i\ge0$. Since $S\otimes_RQ=Q[x_1,\dotsc,x_n]$ is the ring of polynomials in $n$ variables over a field, by the classical Hilbert's syzygy theorem it follows that $\operatorname{Tor}^S_i(M,G)\otimes_RQ$ for all $i>n$. Thus $\operatorname{Tor}^S_i(M,G)$ are torsion $R$-modules for $i>n$.
Let $a\in R$ be a nonzero element. In order to prove that $\operatorname{Tor}^S_{n+1}(M,G)=0$, it suffices to show that (for every $a$) there are no nonzero elements annihilated by $a$ in $\operatorname{Tor}^S_{n+1}(M,G)=0$.
Surely we can assume that $n>0$ (otherwise $R=S$, and there is nothing to prove). Let $0\to\Omega M\to P\to M\to0$ be a short exact sequence of $S$-modules with a projective (or flat) $S$-module $P$ (so $\Omega M$ is "the syzygy module" of $M$). Then we have $\operatorname{Tor}^S_{n+1}(M,G)=\operatorname{Tor}^S_n(\Omega M,G)$.
Both $\Omega M$ and $G$ are $R$-torsionfree modules (and so is the ring $S$). In particular, they are $a$-torsionfree. Consider the left derived tensor products $\Omega M\otimes_S^{\mathbb L}G$ and $(\Omega M/a\Omega M)\otimes_{S/aS}^{\mathbb L}G/aG$, viewed as objects of the derived category of $R$-modules. Then we have a distinguished triangle
$$
\Omega M\otimes_S^{\mathbb L}G\overset{a}\longrightarrow
\Omega M\otimes_S^{\mathbb L}G \longrightarrow
(\Omega M/a\Omega M)\otimes_{S/aS}^{\mathbb L}G/aG \longrightarrow
\Omega M\otimes_S^{\mathbb L}G [1].
$$
From the related long exact sequence of cohomology modules we see that if there are $a$-torsion elements in $\operatorname{Tor}^S_n(\Omega M,G)$ then $\operatorname{Tor}^{S/aS}_{n+1}(\Omega M/a\Omega M,G/aG)\ne0$. This would contradict the assumption of Noetherian induction applied to the ring $R/aR$. End of proof.
EDIT: I was asked to justify the existence of the distinguished triangle. Here is the explanation.
Lemma. Let $S$ be a commutative ring, $a\in S$ be a nonzero-dividing (regular) element, and $M$ and $N$ be two $S$-modules containing no nonzero elements annihilated by $a$. Then there is a distinguished triangle in the derived category of $S$-modules
$$
M\otimes_S^{\mathbb L}N \overset{a}\longrightarrow
M\otimes_S^{\mathbb L}N \longrightarrow
M/aM\otimes_{S/aS}^{\mathbb L}N/aN \longrightarrow
M\otimes_S^{\mathbb L}N [1].
$$
Proof of Lemma. Firstly, there is a distinguished triangle $N\overset a\to N\to N/aN\to N[1]$. Taking the left derived tensor product with $M$ over $S$, we obtain
$$
M\otimes_S^{\mathbb L}N \overset{a}\longrightarrow
M\otimes_S^{\mathbb L}N \longrightarrow
M\otimes_S^{\mathbb L}N/aN \longrightarrow
M\otimes_S^{\mathbb L}N [1].
$$
It remains to construct an isomorphism $M\otimes_S^{\mathbb L}N/aN \simeq M/aM\otimes_{S/aS}^{\mathbb L}N/aN$ in the derived category of $S$-modules. For this purpose, choose a flat resolution of the $S$-module $M$. This is a resolution of an $a$-torsionfree module by $a$-torsionfree modules, so it stays exact after applying the functor $L\longmapsto L/aL$.
Applying this functor to this resolution, we obtain a flat resolution of the $S/aS$-module $M/aM$. Tensoring the former resolution with $N/aN$ over $S$ produces the same complex of $S$-modules (in fact, $S/aS$-modules) as tensoring the latter resolution with $N/aN$ over $S/aS$.
