The model fit of .2 to .4 is from an old paper about aggregate logit models, not respondent-level models - I don't usually see people compute McFadden's rho-squared for HB models, so I'm not sure what my target would be there.
The LL for the null model is easy. If each of your questions has, say 4 alternatives, then a null model with all utilities=0 would choose by chance alone and you'd expect to get 25% correct.
Now follow steps A and B in your post, except use 0.25 instead of RLH and you'll have the null model's log likelihood.
If you have varying numbers of alternatives across sets or a dual response none that contributes to your model then the null LL calculation will be more complicated.