If we try to calculate the oxidation state of nitrogen in $\ce{N2O}$ using the familiar algebraic method, we get oxidation state $+1$ for both nitrogen atoms and that's what I found when I looked it up on the internet.
Well … you get an average oxidation state. This calculation arguably implicitly assumes that all nitrogen atoms be equivalent. In some cases (e.g. hydrazine) they are and the result you get algebraicly is as you would expect from a Lewis depiction. In other cases (e.g. here) this is not the case as the nitrogens are not equivalent (only one is bound to oxygen). Thus, it is obvious that a simple algebraic approach should fail.
But what about the ‘real’ result? What about the resonance structures? Well, this is where things get really difficult. You essentially have two π systems orthogonal to each other and each occupied by four electrons which can manifest as a lone pair on either end and a multiple bond to the other atom. If you really wanted to play the game, you could include another resonance structure as shown below, where all π lone pairs are centred on the terminal nitrogen:
$$\ce{N#\overset{+}{N}-\overset{-}{O} <-> \overset{-}{N}=\overset{+}{N}=O <-> \overset{2-}{N}-\overset{+}{N}#\overset{+}{O}}$$
(It is clear that this third resonance structure contributes least to the overall picture as it has a greater charge separation and charges are separated opposite to what electronegativity would predict.)
The experimental structure shows that the $\ce{N-N}$ distance is slightly shorter than the $\ce{N-O}$ distance which one could use to assume that the $\ce{N-N}$ bond have a slightly larger bond order than the $\ce{N-O}$ bond. But ultimately, they are still very similar (the difference is merely $\pu{4pm}$) so equal bond orders might also be an option. Long story short: short of calculating the electronic distribution (i.e. solving the Schrödinger equation) you won’t be able to arrive at a definite answer for the ‘real’ oxidation states.
So what can you do on paper? Comparing the resonance structures, the leftmost as I have ordered them is slightly better than the central one as the formal charges are distributed according to the different electronegativities. Thus, I would be inclined to give it a slightly higher weight and – in a classroom setting – use it to determine oxidation states. That said, it is too much of an ambiguous example to be seriously used in any examination unless the goal of said examination is to develop the chain of arguments that formed this answer.
