How is the constraint $T\vec{x}\wedge\vec{y}=0$ converted into a set of linear constraints? For $m=2$, one can use the general form of the area $A$ of a parallelogram defined by two vectors $\vec{u}$ and $\vec{v}$: $A=\sqrt{|\vec{u}|^2|\vec{v}|^2-(\vec{u}\cdot\vec{v})^2}$. Substituting $A=0$, this quickly simplifies to $0=u_xv_y-u_yv_x$, which is indeed easily representable in our LP problem. However, it's unclear to me how this extends to higher dimensions. What is a generalized procedure for finding a minimal set of linear constraints that satisfy $T\vec{x}\wedge\vec{y}=0$?

The same way you'd convert any linear function between finite dimensional spaces to a matrix. Choose a basis for the input, i.e. $\vec x$, and a basis for the output, which will be a basis of bivectors, and evaluate the linear function for each basis vector and expand the result in the output basis. This will in a completely mechanical manner give you a matrix. No non-linear operation is involved. The constraints will then be $M\vec x=0$. It's tedious to typeset out the components of $M$ but computing them is a completely mechanical and boring task. If $T:V\to W$ then the expression can be viewed as a linear function $V\to W\wedge W$. If $\mathbf e_i$ is the basis for $V$ and $\mathbf e'_j$ the basis for $W$, then a basis for $W\wedge W$ is $\mathbf e'_j\wedge\mathbf e'_k$ for $j<k$. Then simply distribute $T(\mathbf e_i)\wedge(\sum y_j \mathbf e'_j)$ regrouping into the basis using the antisymmetry of $\wedge$. The coefficients will be the components $i$th column of the matrix.