Determining the structure of N2O using hybridization
Usually you would approach this problem from the other direction. First, find the most likely structure with the correct number of bonds, then deduce the hybridisation. This is also how it works in quantum chemistry: hybridisation is deduced from geometry, not the other way around.

If you try to write the structure of $\ce{N2O}$ you should come up with two equally like structures; these are mesomeric structures as given in the spoiler tag below.


   $$\ce{\overset{-}{N}=\overset{+}{N}=O &lt;-&gt; N#\overset{+}{N}-\overset{-}{O}}$$


From this, we can deduce the most likely hybridisation which will result in a linear molecule. The central atom in a linear molecule typically features two $\mathrm{sp}$ hybrid orbitals and two unhybridised $\mathrm p$ orbitals.

So what about the outer atoms? Well, we cannot say for sure because we do not have enough geometric information. We can tell by the type of bonds that we again need two unhybridised $\mathrm p$ orbitals on the end atoms (remember that the orientations of the double bonds are equivalent!) but we don’t know whether it is a better description for the other two orbitals to be seen as $\mathrm s$ and $\mathrm p$ or two $\mathrm{sp}$ hybrids. In these cases, I would tend to go with ‘unhybridised’ until an experimental or calculative result proves me wrong.



After we have done this, we can indeed look back and realise that we could have determined the structure a priori like this from the general rules; we would just have had to ignore the outer atoms and discuss hybridisation only for the central atom. Indeed, ignoring the hybridisation of terminal atoms gets you very far even in the simplified theory that puts hybridisation first (and is practically wrong).