What is being oxidized? Reduced? What is the oxidizing agent? Reducing agent?

NO3-:

O = -2
N = +5

Zn(s) = 0

Zn(OH)42-:
Zn = +2
O = -2
H = +1

NH3:H = +1
N = -3

N is being reduced and NO3- is the oxidizing agent
Zn is being oxidized and Zn is the reducing agent

There are several approaches to balancing redox reactions in basic solution. Many here do not like the approach I will show you, but it works. You will first balance the reaction as if it occurs in an acidic solution:

1) Divide into two half-reactions:
NO3- --> NH3
Zn --> Zn(OH)42-

2) Balance O by adding H2O:
NO3- --> NH3 + 3 H2O
Zn + 4 H2O --> Zn(OH)42-

3) Balance H by adding H+:
NO3- + 9 H+ --> NH3 + 3 H2O
Zn + 4 H2O --> Zn(OH)42- + 4 H+

4) Balance charge by adding e-:
NO3- + 9 H+ + 8 e- --> NH3 + 3 H2O
Zn + 4 H2O --> Zn(OH)42- + 4 H+ + 2 e-

5) Balance numbers of electrons by multiplying one or both equations by a small number:
NO3- + 9 H+ + 8 e- --> NH3 + 3 H2O
4 Zn + 16 H2O --> 4 Zn(OH)42- + 16 H+ + 8 e-

6. Add the two half-reactions and simplify if needed
4 Zn + NO3- + 13 H2O --> 4 Zn(OH)42- + NH3 + 7 H+

6. Now, to convert this to taking place in basic conditions, add as many OH- to both sides as you have H+. On the side with H+, the H+ and OH- form water that you can then simplify:
4 Zn + NO3- + 13 H2O + 7 OH- --> 4 Zn(OH)42- + NH3 + 7 H2O

4 Zn + NO3- + 6 H2O + 7 OH- --> 4 Zn(OH)42- + NH3 

As long as everything balances (material and charge), you're done.

The oxidation state of each element is shown in the figure below.

NO₃⁻ is reduced because the oxidation number of N decreases from +5 to -3.
Hence, NO₃⁻ is the oxidizing agent.
The half equation is:
NO₃⁻(aq) + 6H₂O(ℓ) + 8e⁻ → NH₃(aq) + 9OH⁻(aq) …… [1]

Zn is oxidized because the oxidation number of Zn increases from 0 to +2.
Hence, Zn is the reducing agent.
The half equation is:
Zn(s) + 4OH⁻(aq) → Zn(OH)₄²⁻(aq) + 2e⁻ …… [2]

[1]×1 + [2]×4, and cancel 8e⁻ and 9OH⁻(aq) on each side. The balanced overall equation is:
NO₃⁻(aq) + 4Zn(s) + 6H₂O(ℓ) + 7OH⁻(aq) → 4Zn(OH)₄²⁻(aq) + NH₃(aq)