Is it necessary to call matrix multiplication $n$ times to find a claw
I think that we can do slightly better than $O(n^{1+\omega})$ for dense graphs, by using rectangular matrix multiplication. A similar idea was used by Eisenbrand and Grandoni ("On the complexity of fixed parameter clique and dominating set", 
Theoretical Computer Science Volume 326(2004) 57–67) for 4-clique detection. 

Let $G=(V,E)$ be the graph in which we want to detect the existence of a claw. 
Let $A$ be the set of (ordered) pairs of vertices in $V$. Consider the following Boolean matrix $M$ of size $|V|\times |A|$: each row is indexed by some 
$u\in V$, each column is indexed by some $(v,w)\in A$, and the corresponding entry 
is one if and only if $\{u,v\}\in E$, $\{v,w\}\in E$ and $\{u,w\}\notin E$. 
Consider the Boolean matrix product of $M$ and its transpose $M^T$. The graph $G$ has a claw if and only if there exists $\{u,v\}\notin E$ 
such that the entry of $MM^T$ in the row indexed by $u$ and the column indexed by $v$ is one. 

The complexity of this algorithm is essentially the complexity of computing the Boolean product of an $n\times n^2$ matrix by a $n^2\times n$ matrix, where $n$ denotes the number of vertices in the graph.
It is known that such a matrix product can be computed in time $O(n^{3.3})$, which is better than $O(n^{1+\omega})$ for the best known upper bound on $\omega$.
Of course,  if $\omega=2$ (as conjectured) then the two approaches give the same complexity $O(n^3)$.