PID's have Krull dimension $1$ (or $0$, if you call a field a PID); $A_Q$ will have Krull dimension $n-1$. So the only option is $n=2$ (the case $n=1$ doesn't apply since $k[x]/x^2$ is not a domain).
However, the $n=2$ case will also not give a PID. The ideal $\langle x_1, x_2, \ldots, x_n \rangle$ is not even locally principal (compute the dimension of the Zariski tangent space).
When I say the title of this question, I thought it would be asking about $k[x,y]/(ax^2+bxy+cy^2+dx+ey+f)$. The answer is that this is a PID if the conic $ax^2+bxy+cy^2+dx+ey+f=0$ is smooth$^*$ and either
(1) $a x^2+b xy + c y^2$ has roots in $\mathbb{P}^1_k$ or
(2) the conic $ax^2+bxy+cy^2+dx+ey+f=0$ has no roots in $k$.
For example, $\mathbb{R}[x,y]/\langle x^2-y \rangle$, $\mathbb{R}[x,y]/\langle xy-1 \rangle$ and $\mathbb{R}[x,y]/\langle x^2+y^2+1 \rangle$ are all PID's, but $\mathbb{R}[x,y]/\langle x^2+y^2-1 \rangle$ is not.
$^*$ In characteristic $2$, I should say regular instead of smooth. For example, I think that $\mathbb{F}_2(t,u)[x,y]/(t x^2+u y^2+1)$ is a PID. I'm not completely confident in this.