Hybridisation of terminal nitrogen in diazomethane
As usual with these small molecules, the bonding situation is a lot more complicated than expected. In this very molecule this is to the extend that not only the Lewis formalism, but also the conventional resonance formalism breaks down.
However, Jan's description is sound, and provides a good starting point for further analysis.
Unfortunately formulating a clear cut valence bond description is difficult, because it involves strongly coupled electrons. This is in part due to the high symmetry of the molecule, $C_\mathrm{2v}$, and the even higher local symmetry of the $\ce{CNN}$ moiety, $C_\infty$. Because of the nature of molecular orbital theory, i.e. being better at describing delocalisation, it is actually easier to understand.

Here are the molecular orbitals computed at the DF-M06L/def2-SVP level of theory:



And here is the same thing in numbers


 Atomic contributions to Alpha molecular orbitals:
 Alpha occ 1 OE=-14.323 is N2-s=1.00
 Alpha occ 2 OE=-14.252 is N1-s=1.00
 Alpha occ 3 OE=-10.099 is C3-s=1.00
 Alpha occ 4 OE=-1.055 is N2-s=0.51 N1-s=0.29 N1-p=0.10
 Alpha occ 5 OE=-0.802 is N2-p=0.31 C3-s=0.31 N1-s=0.14 N2-s=0.12 C3-p=0.07
 Alpha occ 6 OE=-0.518 is C3-p=0.26 C3-s=0.24 H5-s=0.16 H4-s=0.16 N1-s=0.07 N2-s=0.06
 Alpha occ 7 OE=-0.487 is N2-p=0.45 C3-p=0.24 N1-p=0.14 H5-s=0.08 H4-s=0.08
 Alpha occ 8 OE=-0.450 is N2-p=0.61 N1-p=0.29 C3-p=0.09
 Alpha occ 9 OE=-0.407 is N1-s=0.44 N1-p=0.41 N2-p=0.08
 Alpha occ 10 OE=-0.366 is N1-p=0.28 C3-p=0.24 N2-p=0.16 H5-s=0.15 H4-s=0.15
 Alpha occ 11 OE=-0.206 is C3-p=0.57 N1-p=0.38
 Alpha vir 12 OE=-0.082 is N1-p=0.55 N2-p=0.32 H5-s=0.06 H4-s=0.06
 Alpha vir 13 OE=-0.005 is C3-p=0.33 N2-p=0.33 N1-p=0.32


From the visual inspection we can see that the terminal nitrogen has only one orbital that could be classified as a lone pair, i.e mo 9; the hybridisation according to this calculation is approximately sp. There are three bonding &pi; orbitals, i.e. mo 7, 8, 10. And there is also one anti-bonding &pi; orbital, i.e. mo 11. The sigma bond is formed by two bonding (mo 4, 5) and one anti-bonding orbital (mo 6).
This is no coincidence, but pretty much always the case for terminal atoms.

Due to symmetry constraints, the hybridisation pattern at terminal atoms is usually: 


for the orbital forming the single bond $\mathrm{sp}^{\geq1}$
two orthogonal (&pi;) lone pairs $2\times\mathrm{p}$
one (&sigma;) lone pair $\mathrm{sp}^{\leq1}$


Localisation of the orbitals is not easy for this compound. I performed a natural bond orbital analysis, which transforms the (delocalised) canonical orbitals into hybridorbital like ones, but it results in a low occupied carbon lone pair orbital and a partially occupied anti-bonding nitrogen-nitrogen &pi; orbital.


     (Occupancy)   Bond orbital / Coefficients / Hybrids
 ------------------ Lewis ------------------------------------------------------
   1. - 3. core orbitals (omitted)
   4. (1.97583) LP ( 1) N  1            s( 67.37%)p 0.48( 32.60%)d 0.00(  0.03%)
   5. (1.33940) LP ( 1) C  3            s(  0.00%)p 1.00( 99.92%)d 0.00(  0.08%)
   6. (1.99655) BD ( 1) N  1- N  2
               ( 43.82%)   0.6620* N  1 s( 33.02%)p 2.02( 66.72%)d 0.01(  0.26%)
               ( 56.18%)   0.7495* N  2 s( 46.00%)p 1.17( 53.92%)d 0.00(  0.08%)
   7. (1.99587) BD ( 2) N  1- N  2
               ( 51.93%)   0.7206* N  1 s(  0.00%)p 1.00( 99.63%)d 0.00(  0.37%)
               ( 48.07%)   0.6933* N  2 s(  0.00%)p 1.00( 99.80%)d 0.00(  0.20%)
   8. (1.98025) BD ( 3) N  1- N  2
               ( 39.11%)   0.6254* N  1 s(  0.00%)p 1.00( 99.54%)d 0.00(  0.46%)
               ( 60.89%)   0.7803* N  2 s(  0.00%)p 1.00( 99.89%)d 0.00(  0.11%)
   9. (1.99580) BD ( 1) N  2- C  3
               ( 65.65%)   0.8103* N  2 s( 53.76%)p 0.86( 46.18%)d 0.00(  0.05%)
               ( 34.35%)   0.5861* C  3 s( 30.11%)p 2.32( 69.75%)d 0.00(  0.14%)
  10. (1.95419) BD ( 1) C  3- H  4
               ( 60.92%)   0.7805* C  3 s( 34.95%)p 1.86( 65.03%)d 0.00(  0.02%)
               ( 39.08%)   0.6251* H  4 s( 99.91%)p 0.00(  0.09%)
  11. (1.95419) BD ( 1) C  3- H  5
               ( 60.92%)   0.7805* C  3 s( 34.95%)p 1.86( 65.03%)d 0.00(  0.02%)
               ( 39.08%)   0.6251* H  5 s( 99.91%)p 0.00(  0.09%)
 ---------------- non-Lewis ----------------------------------------------------
  12. (0.01076) BD*( 1) N  1- N  2
               ( 56.18%)   0.7495* N  1 s( 33.02%)p 2.02( 66.72%)d 0.01(  0.26%)
               ( 43.82%)  -0.6620* N  2 s( 46.00%)p 1.17( 53.92%)d 0.00(  0.08%)
  13. (0.64416) BD*( 2) N  1- N  2
               ( 48.07%)   0.6933* N  1 s(  0.00%)p 1.00( 99.63%)d 0.00(  0.37%)
               ( 51.93%)  -0.7206* N  2 s(  0.00%)p 1.00( 99.80%)d 0.00(  0.20%)
  14. (0.07754) BD*( 3) N  1- N  2
               ( 60.89%)   0.7803* N  1 s(  0.00%)p 1.00( 99.54%)d 0.00(  0.46%)
               ( 39.11%)  -0.6254* N  2 s(  0.00%)p 1.00( 99.89%)d 0.00(  0.11%)


This is usually an indication of a strongly delocalised system such as a three-centre bond. Investigating the molecule with natural resonance theory gives the rather odd contribution of long range N1 - C3 bond.


 TOPO matrix for the leading resonance structure:

     Atom  1   2   3   4   5
     ---- --- --- --- --- ---
   1.  N   2   2   0   0   0
   2.  N   2   0   2   0   0
   3.  C   0   2   0   1   1
   4.  H   0   0   1   0   0
   5.  H   0   0   1   0   0

         Resonance
    RS   Weight(%)                  Added(Removed)
 ---------------------------------------------------------------------------
    1*     33.02
    2*(2)  32.31    N  1- N  2, ( N  2- C  3), ( N  1),  C  3
    3*(2)  28.81    N  1- C  3, ( N  2- C  3), ( N  1),  N  2
    minor contributions omitted
 ---------------------------------------------------------------------------
          100.00   * Total *                [* = reference structure]


According to NBO, the terminal nitrogen has an $\mathrm{sp}^{0.5}$ &sigma;-type lone pair orbital, two $\mathrm{p}$ &pi;-type bonding orbitals, and a $\mathrm{sp}^2$ &sigma;-type bonding orbital.



Hybridisation is a mathematical tool to analyse a bonding situation. There is no correct way of applying it. It is also very important to understand an atom is never hybridised. (Even though organic chemists tend to use this jargon - even in textbooks.) The orbitals at an atom can be hybridised.