Firstly, as shown in my answer to your `old question', arbitrary smash products of copies of $S/2$ will decompose into wedges of indecomposable spectra of an infinite number of homotopy types (even after suspensions).

At the prime 2, one can use idempotents in $\mathbb Z_2[\Sigma_3]$ to show any spectrum of the form $X 
\wedge X \wedge X$ decomposes as a wedge of the form $Y \vee Y \vee Z$.  

When $X=S/2$, $Y = \Sigma S/2$  (which is likely what you meant) and $Z$ will have the same mod 2 cohomology (as an $A$-module) as $S/\eta \wedge S/2$. (Likely this cohomology module pins down the homotopy type.)