To get dimension 4, I think the norm \begin{equation} \|a\| = \max_{\{i,j\}} |a_i-a_j| \vee \|a\|_\infty, \end{equation} where $a=(a_1,a_2,a_3,a_4)$, works. Again $X$ samples the unit vector basis uniformly.
EDIT: It looks like a variation takes care of dimension 3. Use again \begin{equation} \|a\| = \max_{\{i,j\}} |a_i-a_j| \vee \|a\|_\infty, \end{equation} where $a=(a_1,a_2,a_3)$. This time, let $X$ sample $e_1, e_2, e_3 $ and $-(e_1+e_2+e_3)/2$ uniformly. The first three vectors have norm 1 and the last one norm $1/2$. In the (unlikely) event that my arithmetic is correct, the expectation of $\|X+Y\|$ is $19/16$ and the the expectation of $\|X-Y\|$ is $21/16$.
If this is correct, the only remaining thing is whether the inequality is true for random variables that take on three non zero values.