Let $A$ be a C*-algebra. Consider its cartesian square $A^2$ and define a multiplication on $A^2$ by the identity $$ (x_0,x_1)\cdot (y_0,y_1)=(x_0y_0,x_0y_1+x_1y_0),\qquad x_0,x_1,y_0,y_1\in A $$ This turns $A^2$ into the algebra of "polynomials of one variable of degree 1 with coefficients in $A$": if we denote by $t$ this variable, then each element of $A^2$ can be represented as a polynomial $$ p(t)=x_0+x_1\cdot t,\qquad x_0,x_1\in A, $$ and the multiplication in $A^2$ will be the usual multiplication of polynomials (under the agreement that $t^2=0$): $$ p(t)=x_0+x_1\cdot t,\quad q(t)=y_0+y_1\cdot t $$ $$ \Downarrow $$ $$ p(t)\cdot q(t)=(x_0+x_1\cdot t)\cdot (y_0+y_1\cdot t)=x_0\cdot y_0+(x_0y_1+x_1y_0)\cdot t $$
Is it possibe to define a natural C*-norm on $A^2$?
I mean a C*-norm $\|\cdot\|_{A^2}$ such that $$ \|(x_0,0)\|_{A^2}=\|x_0\|_A,\qquad x_0\in A. $$
Similarly one can define the algebra of polynomials of $m$ variables of degree $n$ with coefficients in $A$, and I am curious in the same question for this case.
"Polynomials" seems like a misleading name; what you're actually doing is trying to adjoin nilpotents.
Anyway, the answer is no, this is already impossible when $A$ is commutative (with no naturality assumption: it is not possible to define any C*-norm on the underlying algebra). The reason is that adjoining nilpotents in the way you want to a commutative C*-algebra gives another commutative C*-algebra, but commutative C*-algebras have no nonzero nilpotent elements by the commutative Gelfand-Naimark theorem.