No, not every sane hypergraph is summable. To see this, let $G$ be the graph obtained from a $7$-cycle $C:=c_1 \dots c_7$ by adding a new vertex $u$ adjacent to all vertices of $C$. Then turn $G$ into a sane hypergraph $H$ by adding the hyperedges $\{c_1,c_3\}$, $\{c_5, c_7\}$, and $\{c_2, c_4, c_6\}$. Towards a contradiction, suppose that $f: V(H) \to \mathbb{Z}_{\geq 0}$ is a function which shows that $H$ is summable. Since $\{u,c_i\}$ is an edge of $H$ for all $i \in [7]$, we must have $f(c_i)=f(c_j)$ for all $i,j \in [7]$. But now the hyperedges $\{c_1, c_3\}$ and $\{c_2, c_4, c_6\}$ have different sums.

No, not every sane hypergraph is summable. To see this, let $G$ be a star with five leaves $\ell_1, \dots, \ell_5$ all adjacent to a vertex $u$. Then turn $G$ into a sane hypergraph $H$ by adding the hyperedges $\{\ell_1,\ell_2\}$ and $\{\ell_3, \ell_4, \ell_5\}$. Towards a contradiction, suppose that $f: V(H) \to \mathbb{Z}_{\geq 0}$ is a function which shows that $H$ is summable. Since $\{u,\ell_i\}$ is an edge of $H$ for all $i \in [5]$, we must have $f(\ell_i)=f(\ell_j)$ for all $i,j \in [5]$. But now the hyperedges $\{\ell_1, \ell_2\}$ and $\{\ell_3, \ell_4, \ell_5\}$ have different sums.