Differential gain of amplifier with current mirror and NO LOAD
Assuming that you are talking about the gain at low frequencies ("DC gain") a derivation of the gain will result in an expression like
$$
Av = g_m (r_{o2} || r_{o3})
$$
where \$g_m\$ is the transconductance of Q2 or Q3 and the \$r_{o2}\$ and \$r_{o3}\$ is the output resistance of these transistors.

The gm of the transistor is easily found to be 
$$
  g_m = \frac{I_C}{V_T} \qquad V_T = \frac{kT}{q}
$$
The output resistance is a property of the transistor and usually described by the Early voltage \$V_A\$. Using the Early voltage the output resistance is
$$
 r_o = \frac{V_A + V_{CE}}{I_C} \approx \frac{V_A}{I_C}
$$
since usually \$V_A \gg V_{CE}\$.
Assuming that both transistors have the same ro the gain is
$$
Av = g_m (r_{o2} || r_{o3}) = \frac{I_C}{V_T} \frac{V_A}{2 I_C} = \frac{1}{2}\frac{V_A}{V_T}
$$
With an Early voltage of 100V and a kT/q of 26mV we get a gain of about 2000.