Counting the nodes of a unidirectional graph
You can build a graph $G'=(V,E')$ that comprises the same set of vertices $V$ from the input graph $G=(V,E)$ and whose set of edges $E'$ includes an edge $(v,u)$ for every edge $(u,v) \in E$. That is, $G'$ is the same graph as $G$, but with the edges reversed; therefore, $G'$ is also topologically sorted (in the same way as $G$, but in the opposite direction). You are looking for the number of successors (+1) of every vertex in $G'$. This can be found using depth-first search, starting a new traversal from every vertex and counting the nodes that you find. The running time of this algorithm is $O(mn)$, where $n$ is the number of vertices of the graph and $m$ is the number of edges. D.W.'s answer provides some useful links with approaches with slightly better running times.