Answer
The ionic compound formed would have the chemical formula $$Li_3N$$
Work Step by Step
The electron configuration of lithium ($Li$) is $$1s^22s^1$$
The electron configuration of nitrogen ($N$) is $$1s^22s^22p^3$$
Both of these atoms would tend to lose, gain or share electrons so that it could reach an octet, that is to have 8 valence electrons, or in other words, to achieve the configuration of a noble gas.
$N$ atom lacks 3 more electrons to reach an octet $(2s^22p^6)$, while $Li$ atom, if loses 1 valence electron in subshell $2s$, would not achieve an octet, but instead achieve the configuration of noble gas $He$, with 2 electrons occupying subshell $1s$, which is also stable.
Therefore, $Li$ atom is willing to lose that 1 valence electron to $N$ atom. However, $N$ atom still needs 2 more electrons to achieve an octet. So, 2 more $Li$ atom would involve in the reaction, each giving away 1 electron to $N$ atom. Finally, all either reach an octet ($N$ atom) or reach a stable state (3 $Li$ atoms).
Eventually, there is 3 $Li$ atoms and 1 $N$ atom involved in the reaction. That means the ionic compound formed would have the chemical formula $$Li_3N$$
