The short answer is no - there is no single inequality criterion.  Already in $\mathbb{R}^3$ everything is much more complicated.  Let me give a sample of inequalities the angles should satisfy.  Denote by $\gamma_{ij}, 1\leq i < j \leq 4$ the six dihedral angles of a Euclidean tetrahedron.  Then:
$$
\gamma_{12}+\gamma_{23} + \gamma_{34}+\gamma_{14} \le  2 \pi
$$
$$
2\pi  \le  \gamma_{12} + \gamma_{13} + \gamma_{14}+\gamma_{23} +  \gamma_{24}+\gamma_{34}  \le  3\pi
$$
$$
0 \le  \cos \gamma_{12} +  \cos\gamma_{13} +  \cos\gamma_{14}+  \cos\gamma_{23} +  \cos\gamma_{24}+ \cos\gamma_{34} \le  2
$$
(See my <a href="http://www.math.ucla.edu/~pak/book.htm">book</a> ex. 42.27 for the proofs of these inequalities - they are not terribly difficult, so you might enjoy proving them yourself). 

This shows that the set of allowed sixtuples of angles is rather complicated (for spherical/hyperbolic tetrahedra with angles close to $\gamma_{ij}$, these angles will have to satisfy these inequalities as well).  The "invariant" you mention corresponds to the unique equation the angles satisfy in the Euclidean space.  The latter is also rather delicate: it is the Gauss-Bonnet equation $\omega_1+...+\omega_4=4\pi$, where $\omega_i$ is the curvature of $i$-th vertex - you need to use spherical cosine theorem to compute it from dihedral angles (see e.g. Prop. 41.3 in my book).  

Finally, you might like to take a look at <a href="http://arxiv1.library.cornell.edu/abs/math/0308239">this</a> interesting paper by Rivin, to see that a similar generalization of the triangle inequality is just as difficult.   To answer your last question (edge lengths from dihedral angles), yes, this is known. I am not an expert on this, but I would start with <a href="http://www.springerlink.com/content/83l8313483871311/">this</a> recent paper.