Consider a strongly connected component $C$ of $G$ and notice that the sought value associated with all vertices $v$ in $C$ is the same. Let's call this value $P(C)$.
Compute all strongly connected components of $G$ (there are several ways to do this in time $O(|V|+|E|)$, see for example Tarjan's algorithm) and let these be $C_1, \dots, C_k$.
Now consider a directed graph $G'=(V', E')$ whose vertex set is $V' = \{ C_1, \dots, C_k\}$ and such that there is a directed edge $(C_i, C_j)$ in $E'$ if and only if $i \neq j$ and $(u,v) \in E$ for some vertex $u$ in $C_i$ and some vertex $v \in C_j$. Notice that $G'$ is a directed acyclic graph, and therefore admits a topological ordering.
Compute such an ordering and assume w.l.o.g. that it is $\langle C_1, \dots, C_k \rangle$ (otherwise we can just rename the components).
This can be done in time $O(|V'|+|E'|) = O(|V|+|E|)$ (actually, Tarjan's algorithm already returns the strongly connected components in reverse topological order).
Notice that if some $C_i$ is a sink in $G'$ (i.e., it has no outgoing edges) then $P(C_i)$ can be immediately computed as $\min_{w \in {C_i}} \text{price}(w)$
If $C_i$ is not a sink, then we have:
$$
P(C_i) = \min \big\{ \min_{w \in C_i} \text{price}(w), \min_{(C_i, C_j) \in E'} P(C_j) \big\}.
$$
By examining the vertices of $G'$ in reverse topological order, we are able to compute all values $P(\cdot)$. Moreover, notice that the time needed to compute $P(C_i)$ is proportional to the number $|C_i|$ of vertices in $C_i$, plus the out-degree $\delta_i$ of $C_i$ in $G'$. Therefore, the overall time complexity of computing $P(\cdot)$ once the topological order is known is upper bounded (up to multiplicative constants) by:
$$
\sum_{i=1}^k ( |C_i| + \delta_i ) = \sum_{i=1}^k |C_i| + \sum_{i=1}^k \delta_i = |V| + |E'| \le |V| + |E|.
$$
