If I'm not mistaken than any rotation can be written in form $e^A$ where $A$ is anti-symmetric matrix.
In 3D for any rotation matrix $R$ we can always find basis that $R = \exp(\omega (e_1e_2^T-e_2e_1^T))$. This does not hold in higher dimensions.
The question is:
Given rotation matrix $R=e^A$, is it possible to find orthogonal basis that $$ A=\omega_{12}(e_1e_2^T-e_2e_1^T) + \omega_{34} (e_3e_4^T-e_4e_3^T) + \dots $$
On
It seems that you are looking for an expansion of an antisymmetric matrix in $n$ dimensions that generalises the result for 3D. For this you need to consider all pairs $i \neq j $ so that:
$$A = \sum_{i \neq j} \omega_{ij} (e_i e_j^T - e_j e_i^T) $$ For the standard orthonormal basis {$e_i $}. There will be some mixed pairs $(i,j)$ such as (1,2) and (2,3), i.e. not all pairs will be disjoint sets.
Apparently you mean by "rotation" in dimension$~n$ any element of the special orthogonal group $\mathbf{SO}(n,\Bbb R)$. Then what you ask for has two parts to it: showing that the exponential map $\mathfrak{so}(n,\Bbb R)\to\mathbf{SO}(n,\Bbb R)$ is surjective (here $\mathfrak{so}(n,\Bbb R)$ is the Lie algebra of real skew-symmetric $n\times n$ matrices), and for a given matrix in $\mathfrak{so}(n,\Bbb R)$ finding an orthonormal basis on which which takes a particularly simple form. The surjectiveness of the exponential map is a fact that you appear to admit; it is true because $\mathbf{SO}(n,\Bbb R)$ is a compact connected Lie group (the proof is not elementary). The existence of a basis adapted to a real skew-symmetric matrix is also true,and can be show using the spectral theory of such matrices.