Consider a triplet of $n$-dimensional ($n > 3$) vectors: $a$, $b$, $c$. Their linear span $span(a, b, c)$ forms a 3-dimensional subspace.
As far as I can judge such a fact is invariant of the coordinate system chosen: looking from the end of vector $c$ calculate the clockwise angle from $a$ to $b$ (with respect only to vector $c$ in this 3D-subspace).
How can this be done with the $n$-dimensional coordinates given?
Orientation can only be given for $n$ vectors in $n$ dimension, by the sign of their determinant.
It has the important property that when moving the $n$ vectors continuously, such that in every moment they are still linearly independent, the orientation remains the same.
In $n+1$ dimension it becomes ambiguous: any $a_1,\dots, a_n$ linearly independent vectors can be smoothly transformed to any other $b_1,\dots, b_n$ linearly independent vectors.
For illustration, consider $(1,0,0)$ and $(0,1,0)$. Using the the $3$rd dimension, we can swap these vectors by a continuous move.
Nevertheless, if we relax the orientation question, we can get the angle $\vartheta$ of two vectors $a, b\in\Bbb R^n$ without need of a 3rd vector, by the usual formula $$\cos\vartheta =\frac{\langle a, b\rangle}{\Vert a\Vert\cdot\Vert b\Vert} $$ where $\langle a, b\rangle=\sum_ia_ib_i$ and $\Vert x\Vert=\sqrt{\langle x, x\rangle} $.