Calculating the zero-point energy for the C2N2 molecule
The adiabatic electronic energy only contains the energy of the electrons. You are missing the energy that the molecule has in form of the vibrational groundstate of the nuclei. This energy can be calculated trivially in the harmonic approximation. Within the harmonic approximation we can define normal modes and each mode corresponds to a 1D harmonic oscillator with its respective normal mode frequency. The eigenenergy of a harmonic oscillator is given by
$$
E_{n} = \hbar \omega(n+1/2), \ n\geq0
$$
Each normal mode of your molecule contributes thus such an energy when you include vibrational energy. The zero point energy refers to the system where each mode is in the groundstate i.e. $n_i=0 $, where $i$ is an index for the normalmode. The zero-point energy  correction is then simply
$$
E_{ZP} = \sum_i^{3N-5} \hbar\omega_i 1/2
$$
where $N$ is the number of atoms and $3N-5$ is the number of normal modes in a linear molecule. In a nonlinear molecule you would have $3N-6$ normal modes. $n$ is the quantum number of the oscillator state.
In your case, you have some degenerate eigenvalues due to symmetry. The $\pi_u$ and $\pi_g$ modes are actually sets of two degenerate modes, which means that their frequency would appear twice in the sum that i denoted. This also works out when we consider your example, we have the following number of modes
$$3\cdot 4 -5=7$$. Since we have to sets of doubly degenerate modes we see only 5 distinct frequencies although there are 7 normal modes.
Plugin all thatt into the formula yields,
$$
E_{ZP} = 1/2(\hbar \omega_1+\hbar \omega_2+\hbar\omega_3 + 2\hbar \omega_4 + 2\hbar \omega_5)
$$
with $\omega_i = 2\pi \nu_i.
$
It's sometimes a bit vague what people mean with zero point energy. Some people use the term for the correction that i defined above, but technically it should include also the electronic adiabatic energy. Which means that you have to add the electronic state energy to the $E_{ZP}$ value.
One might also want to include the contribution due to the rotational states of the molecule, but we can't do that without more information.