It's a fun exercise to derive the conditions of equilibrium here, as this allows us to consider various general combinations of compressible and incompressible materials.
At equilibrium, the total entropy $S$ is maximized: $dS=0$ (and $d^2S<0$).
From the fundamental relation, for container $i$,
$$dS_i=\frac{dU_i}{T_i}+\frac{P_i}{T_i}dV_i.$$
Assume a constant temperature $T$; then,
$$dS=dS_1+dS_2=\frac{dU_1+dU_2}{T}+\frac{P_1-P_2}{T}dV=0$$
because $dV_1=-dV_2$. Since this relation must hold for independent changes in $dV$, we have $P_1=P_2$ at equilibrium. That's familiar enough; pressures generally equalize on either side of a movable partition.
So what is this final pressure? Consider the isothermal bulk modulus $K\equiv -V\left(\frac{\partial P}{\partial V}\right)_T$, which describes the pressure–volume interplay.
From the exchange in volume $dV_1=-dV_2$, we can use the bulk modulus definition to write
$$\frac{V_1}{K_1}dP_1=-\frac{V_2}{K_2}dP_2,$$
which can be integrated to give
$$\int_{P_{1,\text{initial}}}^{P_{\text{final}}}\frac{V_1}{K_1}dP_1=\int_{P_{\text{final}}}^{P_{2,\text{initial}}}\frac{V_2}{K_2}dP_2.$$
Let's first assume both containers contain equal amounts of an ideal gas ($PV=nRT$, $K=P$). Then via integration and algebraic manipulation, we obtain
$$\left[-\frac{nRT}{P_1}\right]_{P_{1,\text{initial}}}^{P_{\text{final}}}=\left[-\frac{nRT}{P_2}\right]_{P_{\text{final}}}^{P_{2,\text{initial}}};$$
$$nRT\left(\frac{1}{P_{1,\text{initial}}}-\frac{1}{P_{\text{final}}}\right)=nRT\left(\frac{1}{P_{\text{final}}}-\frac{1}{P_{2,\text{initial}}}\right);$$
$$P_\text{final}=\frac{2P_{1,\text{initial}}P_{2,\text{initial}}}{P_{1,\text{initial}}+P_{2,\text{initial}}},$$
which is notably not the arithmetic average; the final pressure in your example is not $\frac{3}{2}\,\text{atm}$ but $\frac{4}{3}\,\text{atm}$. (We could have also obtained this from the ideal gas law applied to the final configuration: $P=2nRT/\left(\frac{3}{2}V\right)$.)
Now let's assume container 2 contains a liquid ($V_2$ approximately constant, equivalent to an assumption of $K\gg P$) to obtain
$$nRT\left(\frac{1}{P_{1,\text{initial}}}-\frac{1}{P_{\text{final}}}\right)=\frac{V_2}{K_2}(P_{2,\text{initial}}-P_{\text{final}}),$$
which can be solved for $P_\text{final}$ as
$$P_\text{final}=\frac{\sqrt{(nRT)^2+2nRTP_{1}V_2(2P_{1}-P_{2})/K+(P_{1}P_{2}V_2/K)^2}-nRT+P_{1}P_{2}V_2/K}{2P_{1}V_2/K},$$
where $P_1$ and $P_2$ refer to the initial pressures. For large $K$, a Taylor series expansion of the square-root term to third order gives
$$\sqrt{(nRT)^2+2nRTP_{1}V_2(2P_{1}-P_{2})/K+(P_{1}P_{2}V_2/K)^2}\approx nRT+\frac{P_1V_2(2P_1-P_2)}{K}-\frac{[P_1V_2(2P_1-P_2)]^2}{2nRTK^2},$$
yielding
$$P_\text{final}\approx P_{1}-\frac{P_1V_2(2P_1-P_2)^2}{4nRTK};$$
$$\Delta V\approx \frac{P_\text{final}-P_2}{K}.$$$$\frac{\Delta V}{V_2}= \frac{P_\text{final}-P_2}{K}.$$
For $K\to\infty$, $P_{\text{final}}\to P_{1,\text{initial}}$, as you suspected.
(A nuance that surprised me was the temperature $T$ appearing in the denominator: Shouldn't a higher gas temperature produce more energetic compression of the liquid? Yes, and this is explained by the cubic $P_1$ term appearing in the numerator. For constant $n$ and $V_1$, a greater $T$ must lead to a greater $P_1$.)