The flat dimension of an $R$-module $M$ is the infimum over lengths of flat resolutions of $M$, and the weak dimension (or $\mathrm{Tor}$-dimension) of $R$ is the supremum over all possible flat dimensions of modules. Let's use $\mathrm{w.dim}(R)$ to denote the weak dimension of $R$. As with the global dimension, the weak dimension of $R$ can be computed as the supremum over the set of flat dimensions of the modules $R/I$ for $I$ running over the set of all left-ideals or right-ideals, either is fine!

So, if every ideal is flat, then $\mathrm{w.dim}(R) \leq 1$. What about the converse? If $\mathrm{w.dim}(R) \leq 1$, is it true that every ideal is flat? Let's make a side remark in that if we replace weak dimension with global dimension, and flat with projective, then the answer follows from Schanuel's lemma. However, as far as I know there is no Schanuel's lemma when 'projective' is replaced by 'flat'.

However, we can get away with using part of the proof of Schanuel's lemma. Before continuing, the reader may wish to check out the statement and proof of Schanuel's lemma using a double complex spectral sequence.

So, back to the problem: we know that $\mathrm{w.dim}(R) \leq 1$ and we want to prove that every ideal $I$ of $R$ is flat. Here, $I$ is either a left ideal or a right ideal, but it doesn't matter which. The point is, we have a short exact sequence of $R$-modules $0\to I\to R\to R/I\to 0$. Luckily, in this sequence, $R$ is free, hence projective. We also know that there exists some short exact sequence of the form $0\to F_1\to F_0\to R/I\to 0$ with $F_1$ and $F_0$ flat $R$-modules by assumption on the weak dimension of $R$.

Because $R$ is projective these two short exact sequences fit in a commutative diagram

$$

\begin{matrix} 0 & \to & I & \to & R & \to & R/I & \to & 0\\

~ & ~ & \downarrow & ~ & \downarrow & ~ & \parallel & ~ & ~ \\

0 & \to & F_1 & \to & F_0 & \to & R/I & \to & 0\end{matrix}

$$

As usual, the 'horizontal homology first' spectral sequence or v-h spectral sequence for this double complex shows that the total complex is exact. Thus the sequence

$$

0\to I\to F_1\oplus R\to F_0\oplus R/I\to R/I\to 0

$$

is an exact sequence. The kernel of $F_0\oplus R/I\to R/I$ is just $F_0$. Hence we get a short exact sequence

$$

0\to I\to F_1\oplus R\to F_0\to 0

$$

For any $R$-module $A$, applying $-\otimes_R A$ to this short exact sequence shows that $\mathrm{Tor}_1(I,A) = 0$, so that $I$ is flat.

In the commutative diagram, how are you constructing the map $F_0 \to R$? One can certainly construct a map $R \to F_0$ using the fact that $R$ is projective, but I don't see how you can get a map in the opposite direction instead.

Thanks very much for pointing out this! When typing out this post, I accidentally flipped the diagram and funnily enough the 'proof' still works after this error. I've corrected it now – it should actually be the other way as you pointed out. The essential difference is now the short exact sequence is $0\to I\to F_1\oplus R\to F_0\to 0$, and both $F_1\oplus R$ and $F_0$ are flat, so applying $-\otimes_R A$ to this sequence shows that $\mathrm{Tor}_1(A,I) = 0$ for all $R$-modules $A$ and hence $I$ is flat.

Hopefully that makes more sense.