The compact formula for this sum is
$$
\sum_{i,j,k \in \Bbb Z} {{n}\choose{i+j}}{{n}\choose{j+k}}{{n}\choose{k+i}} = \frac{1}{2} \left( 8^n + \delta_{n,0} \right)
$$
Indeed, let $i+j = m_1$, $i+k = m_2$ and $j+k = m_3$, then
$$
i = \frac{1}{2}\left(m_1+m_2-m_3\right), \quad
j = \frac{1}{2}\left(m_1-m_2+m_3\right), \quad
k = \frac{1}{2}\left(-m_1+m_2+m_3\right)
$$
clearly each $m_k$ is an integer, and must be between $0$ and $n$ for binomials to be non zero. The formulas above maps integers $m_k$ into half-integers, so we need to project those out, using $k \mapsto \frac{1}{2} \left( 1 + \cos(2 \pi k)\right) = \Re\left(\frac{1}{2}\left(1+ \mathrm{e}^{i 2 \pi k}\right)\right)$. Thus
$$
\sum_{i,j,k \in \Bbb Z} {{n}\choose{i+j}}{{n}\choose{j+k}}{{n}\choose{k+i}} = \frac{1}{8} \Re \sum_{m_1=0}^n \sum_{m_2=0}^n\sum_{m_3=0}^n \binom{n}{m_1} \binom{n}{m_2} \binom{n}{m_3}\cdot \\ \cdot \left(1 + \mathrm{e}^{i \pi(m_1+m_2-m_3)} \right) \left(1 + \mathrm{e}^{i \pi(m_1-m_2+m_3)} \right) \left(1 + \mathrm{e}^{i \pi(-m_1+m_2+m_3)} \right)
$$
Now writing out each cosine as a sum of exponentials, and expanding the series we end up with $8$ terms of the form
$$
\sum_{m_1=0}^n \sum_{m_2=0}^n\sum_{m_3=0}^n \binom{n}{m_1} \binom{n}{m_2} \binom{n}{m_3} \mathrm{e}^{i \pi (k_1 m_1 + k_2 m_2 + k_3 m_3)} = \left( \left(1+\mathrm{e}^{i \pi k_1} \right) \left(1+\mathrm{e}^{i \pi k_2} \right) \left(1+\mathrm{e}^{i \pi k_3} \right) \right)^n
$$
Thus
$$
\sum_{i,j,k \in \Bbb Z} {{n}\choose{i+j}}{{n}\choose{j+k}}{{n}\choose{k+i}} = \frac{1}{8} \Re\left( 4 \left(8\right)^n + 4 \delta_{n,0} \right) = \frac{1}{2} \left( 8^n + \delta_{n,0} \right)
$$