Calculate output impedance using negative feedback theory
Yes - its rather simple. From feedback theory we know that the output impedance will be increased by the factor (1+loop gain LG). Therefore:

r(out)=r(ds2)*(1+LG) with 

loop gain LG=(transconductance*load resistance at the source node of M2)

Hence: LG=gm2*[r(ds1)].

Of course, a resistor RD which is connected at the drain of M2 must be considered in parallel to r(out).

Explanation of the feedback effect:

Here we have, in principle, the same situation as for common-emitter amplifiers with Re-stabilization (emitter resistor Re). This resistor provides current-controlled voltage feedback (a rising current Ic causes an increase of the emitter voltage Ve thereby reducing Vbe and counteracting the Ic increase). In the present circuit, the transistor M1 assumes the task of Re and acts as a stabilizing dynamic resistance r(ds1). 

EDIT/Comment:
The above loop gain expression assumes IDEAL feedback conditions, which in reality hardly exists: In this case, ideal voltage feedback (and no current feedback). As a consequence, the loop gain expression contains a certain error. For an exact calculation, both voltage- and current feedback are to be found separately and combined using Middlebrooks formula.