Global minimum at $0$ for $p<2$ is trivial in the sense that you can see it without writing a single equation or inequality (i.e., a non-trivial formula with two sides and a sign in between comparing them).
Observation 1 $e^{-|x|^p}$ is some weighted average of $e^{-ax^2}$ with positive $a$.
Observation 2 The thingy you are interested in is just the variance of $x$ with respect to the probability measure $\mu_y$ whose density is proportional to $e^{-x^2}e^{-|x+y|^p}$. By observation 1, this measure is a mixture of the probability measures $\mu_{a,y}$ with densities proportional to $e^{-x^2}e^{-a(x+y)^2}$. The weight of $\mu_{a,y}$ in that mixture is proportional to something independent of $y$ times $e^{-\frac a{a+1}y^2}$, i.e., when we move $y$ away from the origin, the measures $\mu_{a,y}$ with lower $a$ gain more weight in the composition.
Observation 3. The variance of $x$ with respect to $\mu_{a,y}$ is independent of $y$ and decreases in $a$. To be exact, it is just inversely proportional to $1+a$.
Observation 4. The variance in the mixture is at least the mixture of the variances, which is minimized at $0$ by the independence of individual variances of $y$, observation 3, and the last sentence of observation 2. Also at $y=0$ we have equality because all means are at $0$ by symmetry.
The end.
I wish I could come up with an equally simple argument for $p>2$, but, alas, I don't have one at the moment.