Cannot predict closed-loop gain of this amplifier
We have this circuit 







simulate this circuit &ndash; Schematic created using CircuitLab

First, we need to find the voltage gain for a second stage. 

This gan will be equal to 

$$A_{V2} \approx \frac{R_{C2}||R_L||R_{B2}}{r_{e2}+R_{E2}} \approx \frac{16.3\textrm{k}\Omega}{263\Omega} \approx 62 V/V$$ 

To find the voltage gain for a first stage we need to know the input impedance of a second stage. 

And we can find it using the Miller theorem
How does a Miller cap physically create a pole in circuits?

$$R_{in2} \approx \frac{R_{b2}}{A_{V2}}|| \left(\beta2\cdot (r_{e2}+R_{E2})  \right) \approx 3.36\textrm{k}\Omega $$

Try to derive the full expression for \$R_{in2}\$  

Now the first stage voltage gain: 

$$A_{V1} \approx \frac{R_{C1}||R_{in2}||R_{B1}}{r_{e1}} \approx \frac{2.8\textrm{k}\Omega }{43\Omega} \approx 65 V/V$$

And the input impedance: 

$$R_{in1} \approx \frac{R_{b1}}{A_{V1}}|| \left(\beta1\cdot r_{e1}\right) \approx 2.57\textrm{k}\Omega  $$

So the overall voltage gain is: 

$$A_V =\frac{R_{in1}}{R_g + R_{in1}}\cdot A_{V1}\cdot A_{V2} \approx 1180 V/V $$ 

Do you see your mistake now?  

EDIT

And you can use LTspice to confirm this results. And the easiest will be if you use AC Analysis. And set AC source to 1V. Thanks to this you will get the result directly in V/V. 

See the example

 

As you can see I set the AC source at 1V and the voltage gain of the first stage alone is V(vin2)/V(vin1) equal to 63.4 V/V. 

And by using the AC analysis you can plot Rin, Rout widout any problem. 

For example, the Rin2 is V(vin2)/I(C2)