Do the π-electrons from the C=O bond in tropone contribute to Hückel 's rule?
In addition to Philipp's answer I'd like to share some orbital pictures which shall make the case rather obvious. 

Let me shortly remind you of the key part of the IUPAC definition for aromaticity:


  
  In the traditional sense, 'having a chemistry typified by benzene'.
  


I will therefore follow the same approach as in my answer to "Is phenoxide aromatic?" in comparing the orbital pictures.

The following orbitals are for the π system of benzene, in increasing order of energy (DF-BP86/def2SVP). The two orbitals on the right are degenerate: 



Analogously, the π system for cyclohepta-2,4,6-trien-1-one is extensively delocalised and holds eight electrons. Note that the symmetry of this molecule is only $C_\mathrm{2v}$ removing a lot of constraint from the electronic structure. This is also why the Hückel rule breaks down, since there are no more degeneracies observed.



Also note, that the LUMO of this molecule (pictured below) still has considerable bonding character and that reduction would most likely not destroy aromaticity. (The same statement also applies to oxidation.)