How do you calculate ICMR?
ICMRmax is set by keeping  M5 in saturation. That implies that the maximum voltage at the drain of M5 is \$VDD - V_{dsSatM5}\$. M2 and M3 are PMOS, so their source will be their most positive terminal and their gate voltage must be less than their source voltage if they are to operate properly. That means:

$$
ICMRmax = VDD - V_{dsSatM5} - V_{gsM2}
$$

\$V_{gsM2}\$ is set by the current flowing through M2. Your question implies that the smallest possible value of \$V_{gsM2}\$ is \$V_{t}\$. Strictly, that's true, but since your bias current is already defined and is not zero, \$V_{gsM2}\$ will be larger than \$V_{t}\$. However, for the moment, assuming it is \$V_{t}\$, then ICMRmax = 1.8 - 0.5 - 0.5 = 1V (I assume from your equation that VDD is 1.8V)

ICMRmin is defined by keeping the branch with M2 and M4 in saturation. For this to be true, minimum voltage at the drain of M4 is \$V_{dsSatM4}\$. For M2 to be at the point of saturation 

$$
V_{dsSatM2} = V_{gsM2} - V_{tM2}
$$

ICMRmin will be the gate voltage of M2 under these conditions. This is given by:

$$
ICMRmin = V_{dsSatM4} + V_{dsSatM2} - V_{gsM2}
$$
$$
        = V_{dsSatM4} - V_{tM2}
$$
$$
        = 0.2 - 0.5
$$
$$
        = -0.3V
$$

That makes your ICMR -0.3V to 1V. Actually it will be narrower than that, because \$V_{gs}\$ will be larger than \$V_{t}\$ as mentioned. 

I think that is correct! It's a bit counter-intuitive that the ICMRmin can be negative, but it's a common result for diff amps with PMOS input (you can find it in the textbooks or elsewhere on the web).