Answer to edited question:

It’s still not true for qubit systems. Consider these two unit vectors, both of which are entangled:
$$
|phirangle = frac{1}{sqrt{2}} | 00rangle + frac{1}{sqrt{2}} | 11rangle,\
|psirangle = frac{1}{sqrt{2}} | 01rangle + frac{i}{sqrt{2}} | 10rangle.
$$
Let $rho = |psiranglelangle psi |$, which of course is not on the boundary of the set of separable states; it’s as far away from that set as any state can be. However, we have
$$
text{Tr}Bigl( |phirangle langle phi |^{T_2} rhoBigr) = 0.
$$
To verify this equation, it may help to view the relevant operators as matrices:
$$
|phirangle langle phi |^{T_2} = frac{1}{2}begin{pmatrix}1&0&0&0\0&0&1&0\0&1&0&0\0&0&0&1end{pmatrix}
$$
and
$$
rho = |psirangle langle psi | = frac{1}{2}begin{pmatrix}0&0&0&0\0&1&i&0\0&-i&1&0\0&0&0&0end{pmatrix}.
$$

Original answer:

If I understand it correctly, the statement suggested in the question is false.

For a counter-example, let $rho$ be any entangled PPT state whose partial transpose has an entangled state $|phirangle$ in its kernel. We then obtain
$$
text{Tr}Bigl( |phirangle langle phi |^{T_2} rhoBigr) = text{Tr}Bigl( |phirangle langle phi | rho^{T_2}Bigr) = 0,
$$
but because $rho$ is entangled it cannot be on the boundary of the set of separable states.

Incidentally, no entanglement witness of the form suggested in the question can ever detect entanglement in a PPT state.

To construct an entangled PPT state $rho$ whose partial transpose has an entangled state in its kernel, one may take any unextendable product basis
$$
|phi_1rangle |psi_1rangle,ldots,|phi_mrangle |psi_mrangle in mathcal{A}otimesmathcal{B},
$$
and then let
$$
rho = frac{1}{m} sum_{k = 1}^m |phi_krangle langle phi_k| otimes |psi_krangle langle psi_k|.
$$
The partial transpose looks like this:
$$
rho^{T_2} = frac{1}{m} sum_{k = 1}^m |phi_krangle langle phi_k| otimes |overline{psi_k}rangle langle overline{psi_k}|.
$$
This operator has rank $m and every nonzero vector in its kernel is entangled.