Is a ball a polyhedron?

No a ball is not a polyhedron, even by this definition. In your definition the matrix $A$ is of size $m\times n$, where $m\in\mathbb{N}$ thus the matrix is finite. The integer $m$ is an upper bound on the number of halfspaces which intersect to form the polyhedron.

The reason $m$ is an upper bound is because suppose $A$ has two rows identical. Then there are two hyperspaces which are parallel so at least one of them does not form any part of the polyhedron.

The usual definition of a polyhedron requires that either one intersects a finite number of half-spaces, or one takes the convex hull of a finite set of points.

See the book Convex Polytopes by Branko Grünbaum (either the first or second edition).