Antisymmetric powers of $SO(n)$ representation. I am particularly interested for $SO(3)$.
Let us say that I start with the natural/defining $3$-real-dimensional vectorial representation of $SO(3)$ and I choose the generator of rotation in the $1$-$2$ plane as my Cartan subalgebra. So I have $3$ weight vectors with weights $1$, $0$, $-1$ w.r.t. this generator.
Then is there a way to see what will be the weights under this chosen Cartan of $SO(3)$ of the weight vectors in the $m$-fold antisymmetrization of this $3$-dimensional vectorial representation?


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*It would be great if possible someone can explain the corresponding general result for $SO(n)$, $U(n)$ and $SU(n)$ where too I think the same question can be asked. 


In the particular case of $SO(3)$, one can understand what is happening by general principles.
If $V$ is a $3$-dimensional vector space, then $\wedge^2 V$ is again $3$-dimensional, while $\wedge^3 V$ is one-dimensional, and there is a canonical pairing $V \times \wedge^2 V \to \wedge^3 V$, which induces a natural identification
$\wedge^2 V \cong V^{\vee} \otimes \wedge^3 V$.
Now take $V$ to be the standard $3$-dimensional rep'n of $SO(3)$.  Since an endomorphism of $V$ acts on $\wedge^3 V$ through its determinant, and the elements of $SO(3)$ have determinant $1$ by definition, we see that $\wedge^3 V$ is the trivial rep'n of $SO(3)$.  So we obtain an isomorphism of representations $\wedge^2 V \cong V^{\vee}$.  
But the standard inner product identifies $V$ with $V^{\vee}$, and this identification is compatible with the $SO(3)$-actions (by definition of an orthogonal matrix).
Conclusion:
We have $\wedge^2 V \cong V$ as $SO(3)$-rep's, while $\wedge^3 V$ is the trivial rep'n of $SO(3)$.
