Flat dimension at most n is equivalent to the prescribed higher Tor vanishing

If $F$ is flat, the Tor-one criterion gives ${\mathrm{Tor}}_1^R(N,F)=0$ for every right $N$, and dimension shifting in a projective resolution of $N$ gives ${\mathrm{Tor}}_i^R(N,F)=0$ for all $i>0$.

Suppose $0\to F_n\to \cdots \to F_0\to M\to 0$ is a flat resolution of length $n$. Break it into short exact sequences of successive kernels. The long exact Tor sequence and step 1.1 shift every ${\mathrm{Tor}}_i(N,M)$ with $i>n$ to a positive Tor group of $F_n$, hence to zero.

Conversely, take a projective resolution and put $K_0=M$ and $K_j=\ker (P_{j-1}\to P_{j-2})$ for $j\ge 1$, with $P_{-1}=M$. Repeated dimension shifting identifies ${\mathrm{Tor}}_1^R(N,K_n)$ with ${\mathrm{Tor}}_{n+1}^R(N,M)$; for $n=0$ this is the identity. The assumed vanishing makes $K_n$ flat by the Tor-one criterion. Truncating at $K_n$ gives a length-$n$ flat resolution, including the case $n=0$.