Here goes $n=2$ a la Fedor Petrov.
Notice that his argument is based on the smart identities $x_z+y_z=|z|$, $z_x=|x|+y_x$ and $z_y=|y|+x_y$ where $x,y$ are 2 vectors in $\mathbb R^2$, $z=x+y$ and $x_y$ is the (signed in general, but in our setting everything is non-negative) length of the projection of $x$ to $y$. Fedor's idea can be made into a one-liner: consider $\Phi_1(x,y)=|x|+|y|-x_y-y_x$, check the identity $\Phi_1(x,y)-\Phi_1(z,x)-\Phi_1(z,y)=|x|+|y|-|z|$ and telescope.
For $n=2$ the right function to consider is $$ \Phi_2(x,y)=2|x||y|-|x|y_x-|y|x_y $$ I wrote it in the symmetric form, but actually $|x|y_x=|y|x_y=|x||y|\cos\theta$ where $\theta$ is the angle between $x$ and $y$. The difficulty is that in this case the decay "at infinity" is not fast enough to send the sum of faraway terms to $0$: the typical term there is $2|x||y|(1-\cos\theta)=(2|x||y|\sin\frac{\theta }2)2\sin\frac{\theta}2\approx A\theta$ where $A$ is the (invariant) area of the parallelogram spanned by $x$ and $y$. The sum of these expressions at infinity adds an extra term $A$ times the angle between $x$ and $y$ with $-$ sign (the precision is now good enough to ignore the rest; like Fedor, I'll leave the "routine convergence checks" to the reader). So, for two perpendicular unit vectors, we get $2\cdot1\cdot1-0-0-1\cdot\frac \pi 2$ as requested.
I wonder if we can continue with that a bit. The first task would be to find $\Phi_3$ but, even if we are lucky and it exists, it may require even more correcting terms when trying to telescope and we can get a transcendental problem there.
