I am not sure how helpful this may be, but it is too long for a comment.

There may be no reference where people carry out the computation using the usual Hochschild complex because *no one* wants to use that complex, in general, to compute the Hochschild cohomology groups of an algebra, when better resolutions exist.

For free algebras (or more generally quiver path algebras with no relations over a semi-simple base) there is a canonical bimodule resolution of $A=TV$ of the form (starting at degree $-1$)

$$0\to A\otimes V\otimes A\to A\otimes A\to A \to 0$$

where the leftmost map is $d(x\otimes v\otimes y) = xv\otimes y - x\otimes vy$ and the second map is the product.

Let me call this complex $L$. Note that there is an obvious inclusion $\iota:L\to B$ where $B$ is the double sided bar construction of $A$, and this inclusion is a chain map. In the other direction, there is a map $\pi : B\to L$ that is the identity in degree $-1$ and $0$, on $A^{\otimes 3}$ acts as follows:

$$
\pi(f\otimes v_1\cdots v_n\otimes g) = \sum fv_1\cdots v_j\otimes v_{j+1}\otimes v_{j+2}\cdots v_n g\in A\otimes V\otimes A,
$$
while it is necessarily zero in the other remaining degrees. You can check that $\pi$ is a chain map, and $\pi \iota=1$. It is a bit more tricky to check that $\iota\pi$ is chain homotopic to $1$. In the end, what you get is that $\hom_{A^e}(B,A)$ and $\hom_{A^e}(L,A)$ are chain homotopy equivalent.

Of course $\hom_{A^e}(L,A)$ is naturally isomorphic to
$$0\to A\to \hom(V,A) \to 0$$
where the only non-zero map sends $a\in A$ to $V\to A$ such that $v\longmapsto [a,v]$, giving in your case $\mathrm{HH}^0(A) = \mathrm{HH}^1(A) = A$ as $V$ is one dimensional. What you'd have to do now is figure out how to lift these obvious cycles to representative cycles in $\hom_{A^e}(B,A)$, but you can do this precomposing with the chain map $\pi$.

whydo you want to carry out the computation with the Hochschild complex? $\endgroup$