From Gelfand-Neumark Theorem, we know that topological properties of a compact Hausdorff space $X$ are encoded in the abelian $C^*$-algebra of continuous complex-valued functions on $X$ (with $||f||= \sup |f|$). For example, the existence of non-trivial idempotents would imply that the space $X$ is not connected.
Is there such an algebraic description to determine whether $X$ is simply connected or not ?
I am not really an expert in non-commutative point-set/algebraic topology (but I used to think about it a decent amount). So take my answer with that in mind.
So the silly answer is yes and the more interesting answer is no.
Specifically, The Gelfand correspondence sets up a contravariant equivalence of categories between compact Hausdorff space with continuous maps and unital abelian $C^*$-algebras with $*$-homomorphisms. So through this you can unwind (wind up?) the definition of simply connected though this equivalence to get a characterization in terms of the associated $C^*$-algebras. So it would be something like $A$ is simply connected if every map $\phi:A\rightarrow C(S^1)$ is homotopic to the constant maps. Where $C(S^1)$ is continuous functions on the circle and note that homotopy does translate well into the $C^*$-algebraic framework.
However, this isn't particularly useful. In fact, what I think that you are asking is "is there a characterization which is useful for non-abelian $C^*$-algebras". This is the case with connectedness that you give above.
Here I think the answer is no and, in fact I will say something stronger. I think that there can be no such characterization (Here I'm being a little bolder). The reason is because of the contravariance of the equivalence. In particular, any topological invariant that comes from maps of some space $X$ (in the above case $S^1$) to the desired space $Y$ will not bear anything fruitful for general $C^*$-algebras. Because on the level of the algebras you get a map from $C(Y)\rightarrow C(X)$. So in order for this to work for any $C^*$-algebras it would in particular have to work for the non-abelian simple ones (these are largely the ones of interest). And they can't have any morphisms into abelian algebras.