Let $K$ be the field of Hahn series in an indeterminate $t$, with exponents in $\mathbb{R}$ and coefficients in $\mathbb{F}_2$, and where we denote its valuation by $\nu$.
Define $$A:=\{a\in K:\nu(a)\geq 0\},$$ and for each $q\in\mathbb{R}$, set $I_q:=t^qA$ and $I_{>q}:=\bigcup_{r>q}I_r$. Also set $P:=A/I_{>1}$.
I know that an $A$-module $M$ is flat iff $\text{ann}(t,M)=0$, and therefore all $A$-submodules of $K$ are flat. I suspect that if $I_{>1}\leq L\leq A$ is an $A$-submodule of $K$, then $L/I_{>1}$ is a flat $P$-module. (We can identify the $P$-modules with the $A$-modules $M$ satisfying $I_{>1}M=0$.)
How might I go about checking this?
$L/I_{>1}$ is flat iff $L/I_{>1}\otimes(-):\text{Mod}_P\to\text{Mod}_P$ is an exact functor iff whenever $\varphi:M\to N$ is a $P$-monomorphism, we have that $\text{id}_{L/I_{>1}}\otimes\varphi$ is also a $P$-monomorphism.
If $\varphi:M\to N$ is a $P$-monomorphism, then it is an $A$-monomorphism, so that $\text{id}_L\otimes\varphi:L\otimes M\to L\otimes N$ is an $A$-monomorphism. But I don't know where to go from here...
This is not true. For instance, $I_{1/2}/I_{>1}\cong A/I_{>1/2}$ so $I_{1/2}/I_{>1}\otimes A/I_{>1/2}\cong A/I_{>1/2}$ as well. This means that if you take the inclusion $I_{1/2}/I_{>1}\to P$ (which sends the generator to $t^{1/2}$) and tensor it with $A/I_{>1/2}$ you get a map $A/I_{>1/2}\to A/I_{>1/2}$ which sends the generator to $t^{1/2}$ which is thus not injective.
(The moral here is that when you mod out $I_{>1}$, modules that were nice and torsion-free can gain torsion that makes them not flat. A much simpler analogous example is that $2\mathbb{Z}$ is flat over $\mathbb{Z}$ but $2\mathbb{Z}/4\mathbb{Z}$ is not flat over $\mathbb{Z}/4\mathbb{Z}$.)