I have to calculate
$$\sum_{1 \le i <j <k \le n}^{} \arctan\left(\dfrac{n}{i} \right) \arctan\left(\dfrac{n}{j} \right)\arctan\left(\dfrac{n}{k} \right) $$
We notice for any positive number $x$, we have
$$\tan^{-1}(x) = \frac{1}{2i}\log\left(\frac{1+ix}{1-ix}\right) = \Im\log(1 + ix)$$
We can rewrite the sum at hand as
$$\sum_{k=1}^\infty \tan^{-1}\frac{n}{k} = \Im\left[ \sum_{k=1}^\infty \log\left(1 + \frac{n}{k}\right) \right] $$
but I can;t go on.
About the asymptotic behaviour: we may replace $\arctan\frac{n}{a}$ with $\frac{\pi}{2}-\arctan\frac{a}{n}$ in order to only deal with values of the arctangent function over the interval $I=[0,1]$. Over such interval $\arctan(x)=x-\Theta(x^3)$, hence the main term of the asymptotic behaviour is the same as the main term of $$ \frac{\pi^3}{8}\sum_{1\leq i<j<k\leq n}\left(1-\frac{2i}{\pi n}\right)\left(1-\frac{2j}{\pi n}\right)\left(1-\frac{2k}{\pi n}\right)$$ which can be easily evaluated (in explicit terms) through Stirling numbers of the first kind.