Sasha's argument is pretty technological. You can really do this almost by hand, though.
Let $A$ be the algebra in question, let $r$ be its Jacobson radical (that is, the subspace of strictly upper triangular matrices), and let $E$ be the subalgebra of the diagonal matrices in $A$ (which is a complement to $r$) Notice that both $A$ and $r$ are $E$-bimodules.
The algebra $A$ has a projective resolution as a bimodule of the form $A\otimes_E r^{\otimes_E *}\otimes_E A$ which looks exactly like the Hochschild resolution but the inner copies of $A$ have been replaced by $r$, and all tensor products involved are over $E$ and not over the base field; the differentials in the complex have exactly the same formula as the usual Hochschild differential. This can be checked easily —it is a nice exercise— but you can find the details in a nice paper by Claude Cibils on square-zero algebras, if I recall correctly.
Now, $HH_*(A)$ is $Tor^{A^e}_*(A,A)$, so it can be computed as the homology of the complex obtained from $A\otimes_E r^{\otimes_E *}\otimes_E A$ by applying the functor $A\otimes_{A^e}(\mathord-)$. You should explicitly describe this complex: its only non-zero term is the $0$th one, so its homology is very, very easy to compute!
(The same thing can be done for every triangular algebra, that is, every algebra whose ordinary quiver is acyclic)