Define $g(x)=f(x)-f(0)$. Then, setting $x_2=...=x_n=0$ you see $f(x/n)=f(x)/n+f(0)-f(0)/n$ or equivalently $g(x/n)=g(x)/n$. This implies $g(x+y)=2(g(x+y)/2)=2g((x+y)/2)=2(g(x)+g(y))/2=g(x)+g(y)$.
Furthermore, we obtain $g(m/n x)=g(mx)/n=m/n (g(mx)/m)=m/n g(x) $ for all $m,n \in \Bbb N$. So $g(qx)=qg(x)$ for every positive rational $q$. Indeed, this holds for any rational $q$ since $g(-x)+g(x)=g(0)=f(0)-f(0)=0$, which implies $g(-qx)=-g(qx)=-qg(x)$.
In general, $g$ can be any linear map of $\Bbb R$ to $\Bbb R$, where $\Bbb R$ is viewed as a $\Bbb Q$ vector space, but if you demand that $f$ is continuous, $g(q)=qg(1)=cq$ implies $g(r)=cr$ for every real $r$. So $f(x)=cx+f(0)=cx+b$ are the only continuous solutions.