Is it a valid graph canonical form?
A canonical form of a graph is a mapping $C$ from the set of graphs to the set of graphs such that $C(G)$ is isomorphic to $G$, and $C(G) = C(H)$ iff $G$ is isomorphic to $H$. Your function doesn't map graphs to graphs, so it's not a canonical form.

A more relevant notion is a (complete) invariant. This is a mapping $I$ from the set of graphs to some other set such that $I(G) = I(H)$ iff $G$ is isomorphic to $H$. Your mapping, as stated, is not a graph invariant. For example, consider the graph consisting of a triangle plus an isolated vertex. Consider the graph $G$ in which the triangle vertices are $1,2,3$, and the graph $H$ in which the triangle vertices are $1,2,4$. The two graphs are isomorphic, but the vector $\mathit{Triangle}(\cdot)$ is different.

We can try to fix this problem by sorting the vector. This will ensure that the vectors corresponding to $G,H$ above are the same. However, consider $H'$, which consists of a triangle connected by an edge to a vertex. The graphs $H$ and $H'$ have the same sorted triangle vector, but are not isomorphic. So your suggestion is not even a complete invariant.

Finally, you ask how easy it is to calculate your function. There is a trivial $O(n^3)$ algorithm which goes over all triples of vertices, and this can be improved to $O(n^\omega)$ using fast matrix multiplication: if $A$ is the adjacency matrix of the graph, then $(A^3/2)_{vv}$ is the number of triangles containing $v$.