Since a lie group is a manifold with the structure of a continuous group, then each point of the manifold has some scalar curvature R. Is there a nice formula which relates the lie algebra of the group to the scalar curvature at a point of the manifold?
See Exercice 1 in Chapter 4 of Do Carmo's "Riemannian Geometry".
The formula is $R(X,Y)Z = \frac 1 4 [[X,Y], Z]$.
In particular, if $X$ and $Y$ are orthonormal, the sectional curvature of the generated plane is
$K(\sigma)= \frac 1 4 \|[X,Y]\|^2$
Which is always $\geq 0$.
EDIT: In view of the comments, it is important to add that this is for a bi-invariant metric.
For left-invariant (or right-invariant) metrics, this paper of Arnold gives a formula for the sectional and Riemannian curvatures, in terms of the adjoint of the Lie bracket operation in the metric.
One result which I think will be what you are interested in is this,
(corrected and clarified in response to Jose's pointers)
This mapping of the connection in terms of the Lie Algebra can be fruitfully used to achieve simpler expressions for various other quantities, like most beautifully the statement that the scalar curvature becomes one-fourth of the dimension of the Lie Group!