I saw this question as a bonus from a past exam, and here's my solution for verification.
I argued like so. I said suppose there is such a surjective homomorphism $f$, then $f(0)=(0,0,0)$, $f(1)= (1,1,1)$ by ring homomorphism axioms. Suppose now that $f(x)= (a,b,c)$, where $a,b,c$ are either $0$ or $1$. Then $f(x^2)= (a^2,b^2,c^2)= (a,b,c)$ and same thing for $f(x^n)$ for any $n\geq 1$. Now this implies that any $p(x)$ will be mapped to either $(a,b,c)$ or $(a+1,b+1,c+1)$, depending on if they have a constant term (1). This means that $(a+1,b,c)$ for example is not in the image of $f$. Done.
Is this a good argument? Thanks in advance.