Yes; if $f:=f_1(x_1)+f_2(x_2)$ is irreducible in $\Bbb{C}[x_1,x_2]$ then it is prime, so the quotient $\Bbb{C}[x_1,x_2]/(f)$ is an integral domain. For every $r\geq3$ you have
$$\Bbb{C}[x_1,\ldots,x_r]/(f)\cong(\Bbb{C}[x_1,x_2]/(f))[x_3,\ldots,x_r],$$
which is again an integral domain, so $f$ is also prime and hence irreducible in $\Bbb{C}[x_1,\ldots,x_r]$.
Alternatively, suppose $f$ is reducible in $\Bbb{C}[x_1,\ldots,x_r]$ for some $r\geq3$, and write $f=gh$ with $g,h\in\Bbb{C}[x_1,\ldots,x_r]$ nonconstant. Because $f$ is irreducible in $\Bbb{C}[x_1,x_2]$ either $g$ or $h$ must have a monomial term that is a multiple of some $x_k$ for some $k\geq3$. That is to say, there exists some $k\geq3$ such that either $\deg_{x_k}g>0$ or $\deg_{x_k}h>0$. Then
$$\deg_{x_k}f=\deg_{x_k}gh=\deg_{x_k}g+\deg_{x_k}h>0,$$
a contradiction. So $f$ is also irreducible in $\Bbb{C}[x_1,\ldots,x_r]$.