Your proposed structure is wrong. Nitrogen does not exceed the octet in any of its known compounds (and even if $\ce{NF5}$ will be discovered it will not exceed the octet according to everything we know now). However, if you have a formal negative charge that means an additional electron added to the 5 nitrogen usually has; if four of those six electrons are used to build the double bonds, there is still a lone pair on nitrogen for a total of 10 electrons.
If you are having trouble determining Lewis structures, there are four quick calculations that you can perform to help you:
Add up all valence electrons the atoms are bringing into the compound.
Each nitrogen has five electrons plus there is one negative charge (additional electron) so:
$$3\times5+1=16\tag{1}$$
Add up how many valence electrons would be needed so that each atom has an octet (for hydrogen: dublet) of its own.
Each nitrogen would want eight electrons so:
$$3\times8=24\tag{2}$$
Take $(2)-(1)$. This represents the number of electrons the atoms must share, i.e. the number of bonds.
$$24-16=8\tag{3}$$
Take $(1)-(3)$. This represents the number of electrons that do not have to take part in bonds; these must then be distributed as lone pairs.
$$16-8=8\tag{4}$$
Then, start drawing but make sure you have as many lone pairs and bonding electrons as the equations state. Ignoring the lone pairs, we can get the following possible structures for $\ce{N3-}$:
$$\ce{N#N-N}\qquad\qquad\ce{N=N=N}\qquad\qquad\ce{N-N#N}$$
(The exercise of distributing four lone pairs across the three nitrogens so that each ultimately has eight valence electrons is left to the reader because I am too lazy to open ChemDraw to draw the structures.)
After you have done that, you need to take a look at potential formal charges. For that, split each bond homogenously (i.e. give each atom one of the bonding electrons) and count. Compare that count to what an atom should have; the difference corresponds to the atom’s formal charge. (Because electrons are negative, an additional electron corresponds to a charge of $-1$.) When done for those three structures, we arrive at:
$$\ce{N#\overset{+}{N}-\overset{2-}{N}}\qquad\qquad\ce{\overset{-}{N}=\overset{+}{N}=\overset{-}{N}}\qquad\qquad\ce{\overset{2-}{N}-\overset{+}{N}#N}$$
In each of those cases, the formal charges sum up to the overall charge of the molecular ion ($-1$) which is an indication that we have done it correctly. (Again, I have sheepishly left out the lone pairs; you can use my formal charges to determine where they should have been and how many.)
There is no principle of zero formal charges. However, when debating between different structures, a structure with less formal charges is often (not always!) more ‘favourable’. (The actual term should be ‘contributes more to the overall picture’ but that may confuse too much at this stage.)
But which of the three is correct? They all are! In fact, this is what is known as mesomery: we have a number of (resonance) structures that all explain the actual compound a little bit but neither holds the absolute truth. To show this, resonance arrows are usually drawn between the depictions:
$$\ce{N#\overset{+}{N}-\overset{2-}{N} <-> \overset{-}{N}=\overset{+}{N}=\overset{-}{N} <-> \overset{2-}{N}-\overset{+}{N}#N}$$
The key difference between correct structures and your proposition is that the central nitrogen atom can never carry a negative formal charge as it needs to accomodate four bonds to its neighbours which is only possible for $\ce{N+}$.
As for the answer given in the homework response: It is not strictly correct because it is incomplete. All three structures should be marked as correct – until the concept of resonance has been formally introduced at which point only a combination of the three should be.
