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(I write $HH^\bullet(\Lambda)$ what you write $H^\bullet(\Lambda,\Lambda)$. When I become Emperor of Notation, everyone will!)
Since $A_n\cong A_1\otimes\cdots\otimes A_1$, the Künneth formula for Hochschild cohomology (which is proved in Cartan-Eilenberg, Theorem XI.3.1, for example) tells you that $HH^\bullet(A_n)\cong HH^\bullet(A_1)^{\otimes n}$. It is enough, then, to compute $HH^\bullet(A_1)$.
The computation of the whole of $HH^\bullet(A_1)$ is not difficult to carry out directly.
You can climb on the shoulders of others and do the following, too. First, it is easy to check that the center of $A_1$ is $k$, so that $HH^0(A_1)=k$. Second, Jacques Dixmier shows, in his Algèbres enveloppentes, that every derivation of $A_1$ is inner, so that $HH^1(A_1)=0$. Finally, the algebra $A_1$ is a Calabi-Yau algebra of global dimension $2$, so in particular it satisfied van den Bergh duality and $HH^2(A_1)=HH_0(A_1)$. The latter vector space is $A_1/[A_1,A_1]$, and a pleasurable computation shows this is zero.
Alternatively, in view of Calabi-Yau-ness of $A_n$, now of global dimension $2n$, we have that $HH^\bullet(A_n)=HH_{2n-\bullet}(A_n)$, and a theorem of Wodzsicki Wodzicki tells us that $HH_\bullet{2n-\bullet}(A_n)$ HH_{2n-\bullet}(A_n)$is isomorphic tothe to the algebraic de Rham cohomology of$n$-dimensional affine space$\mathbb A^n$, which is just$k$. This way we get$HH^\bullet(A_n)=k$in one big swoop. 1 (I write$HH^\bullet(\Lambda)$what you write$H^\bullet(\Lambda,\Lambda)$. When I become Emperor of Notation, everyone will!) Since$A_n\cong A_1\otimes\cdots\otimes A_1$, the Künneth formula for Hochschild cohomology (which is proved in Cartan-Eilenberg, Theorem XI.3.1, for example) tells you that$HH^\bullet(A_n)\cong HH^\bullet(A_1)^{\otimes n}$. It is enough, then, to compute$HH^\bullet(A_1)$. The computation of the whole of$HH^\bullet(A_1)$is not difficult to carry out directly. You can climb on the shoulders of others and do the following, too. First, it is easy to check that the center of$A_1$is$k$, so that$HH^0(A_1)=k$. Second, Jacques Dixmier shows, in his Algèbres enveloppentes, that every derivation of$A_1$is inner, so that$HH^1(A_1)=0$. Finally, the algebra$A_1$is a Calabi-Yau algebra of global dimension$2$, so in particular it satisfied van den Bergh duality and$HH^2(A_1)=HH_0(A_1)$. The latter vector space is$A_1/[A_1,A_1]$, and a pleasurable computation shows this is zero. Alternatively, in view of Calabi-Yau-ness of$A_n$, now of global dimension$2n$, we have that$HH^\bullet(A_n)=HH_{2n-\bullet}(A_n)$, and a theorem of Wodzsicki tells us that$HH_\bullet{2n-\bullet}(A_n)$is isomorphic tothe algebraic de Rham cohomology of$n$-dimensional affine space$\mathbb A^n$, which is just$k$. This way we get$HH^\bullet(A_n)=k\$ in one big swoop.