It seems clear to me that blocks should have something to do with the decomposition of the category as a direct product of subcategories. A decomposition into a product of two factors corresponds exactly to an idempotent in the center of the category (recall that the center of an abelian category is the ring of natural transformations of the identity, it equals the center of a ring when the category is the category of all modules over the ring). Hence, one could define a block to be such an idempotent and a primitive block to be a primitive idempotent. Thus a block is a subcategory that is a direct factor and a primitive block is an indecomposable such direct factor.
I think that if one wants a completely general definition this may be the only way to go as there are situations where there are lots of idempotents but no primitive idempotents. (Consider for instance modules over the Boolean algebra of subsets of an infinite set modulo the finite subsets.) In particular I don't think that the definition quoted by Noah would be suitable in the case when there are no indecomposables. (If every object is a sum of indecomposables I think the definition gives what I propose to call primitive blocks.)
A comment on the case of Lie algebra representations. The enveloping algebra of a Lie algebra of course contains no non-trivial idempotents and thus neither does its center. What one does however is to look at various subcategories where (some?) elements of the center have generalised eigenspace decompositions. This introduces idempotents in the centre of the category (which actually comes from idempotents in some suitable completion of the center of the enveloping algebra).
[Added] When every object in the category has finite length blocks are in bijection with subclasses $S$ of the class of simple objects closed under the relation of having non-trivial extensions (in either direction). Indeed, the only non-trivial part is to show that any object is the direct sum of one object all of whose Jordan-Hölder factors are in $S$ and another one none of whose Jordan-Hölder factors are in $S$. If $M$ is an object all of whose Jordan-Hölder factors are in $S$ and $P$ is a simple module not in $S$, then all extensions of $M$ by $S$ are trivial (as is shown by induction over the length of $M$) and the same for $M$ having no Jordan-Hölder factor in $S$ and $P$ being in $S$. The splitting of an object as such a direct sum is now done by induction over the length of the object.