I think I've found the answer for the single player case:
If we write $e_{i}$ for the expected remaining length of the game if $i$ cards are facedown, then we can work out that:
(i). $e_{5} = \frac{1}{6}(1) + \frac{5}{6}(e_{4} + 1)$
(ii). $e_{4} = \frac{2}{6}(e_{5} + 1) + \frac{4}{6}(e_{3} + 1)$
(iii). $e_{3} = \frac{3}{6}(e_{4} + 1) + \frac{3}{6}(e_{2} + 1)$
(iv). $e_{2} = \frac{4}{6}(e_{3} + 1) + \frac{2}{6}(e_{1} + 1)$
(v). $e_{1} = \frac{5}{6}(e_{2} + 1) + \frac{1}{6}(e_{0} + 1)$
(vi). $e_{0} = \frac{6}{6}(e_{1} + 1)$
(vi) and (v) then give us (vii). $e_{1} = e_{2} + \frac{7}{5}$;
(vii) and (iv) then give us (viii). $e_{2} = e_{3} + \frac{11}{5}$;
(viii) and (iii) then give us (ix). $e_{3} = e_{4} + \frac{21}{5}$;
(ix) and (ii) then give us (x). $e_{4} = e_{5} + \frac{57}{5}$;
(x) and (i) then give us $e_{5} = 63 $
We can then add up to get $e_{0} = 63 + \frac{57}{5} + \frac{21}{5} + \frac{11}{5} + \frac{7}{5} + 1 \approx 83.2$$e_{0} = 63 + \frac{57}{5} + \frac{21}{5} + \frac{11}{5} + \frac{7}{5} + 1 = 83.2$.
Now, how would one generalize this to find the expected length of game with $n$ players?