What the RL as view by this circuit
Let's look at the 2nd stage.
The first thing to do is to work out the DC quiescent point for your 2nd stage. Assuming \$\beta\approx 200\$, I get \$I_{\text{B}_2}=\frac{V_{_\text{TH}}-V_\text{BE}}{R_{_\text{TH}}-\beta\cdot R_\text{E}}\approx 5.5\:\mu\text{A}\$ and therefore \$I_{\text{E}_2}\approx 1.1\:\text{mA}\$. This means \$r_e^{'}\approx 24\:\Omega\$. And if your frequency is \$1\:\text{kHz}\$, then \$X_\text{C}\approx 5.3\:\Omega\$. Reflected back to the base I get about \$5.9\:\text{k}\Omega\$. So the total load on the first stage is about \$5.9\:\text{k}\Omega\mid\mid 26\:\text{k}\Omega\mid\mid 2.9\:\text{k}\Omega\$ or about \$\approx 1.81\:\text{k}\Omega\$. This means that only about 27.4% of the 1st stage voltage gain is retained. About \$27.4\%\cdot 10.67\approx 2.92\$, or so.
Very close to what you measured with simulation.