Pathological product space norm

Let $X=Y=\mathbb R$ with the absolute value norm and define $n(a,b)=\sqrt{2a^2+2b^2-3ab}$. This is a norm on $\mathbb R^2$ because it is the quadratic form of the positive definite matrix $A=\left( \begin{smallmatrix} 2 & -3/2 \\ -3/2 & 2 \end{smallmatrix}\right)$.

Then $N(v) = n(|v_1|,|v_2|)$ is not a norm because the triangle inequality fails for $u=(1,1)$, $v=(1,-1)$: we have that $N(u)=N(v)=1$, but $N(u+v)=2\sqrt{2}$.

This is only an illustration to Christian Remling's beautiful answer; here are the concentric "balls" around the origin for this "norm":

This is a comment rather than an answer but I am not entitled. The missing requirement on the two dimensional norm for the statement to be valid is that it be increasing in the natural sense that if $|x|\leq|x_1|, |y|\leq|y_1|$, then $n(x,y)\leq n(x_1,y_1)$. Norms without this property can easily be obtained by rotating a non-circular ellipse in standard position (as in the above answer).