Torsion pairs and projective dimension

In this answer I freely use results and notation from the book you refer to. Let me know if I need to fill in more details.

Let $U=\operatorname{Ext}^1_A(T,N)$. We can assume $\operatorname{pd}_B (U)=n>2$ since the statement holds trivially for smaller values. Let $$0 \rightarrow \Omega_B U \rightarrow P_B \rightarrow U \rightarrow 0$$ be the first step in a projective $B$-resolution of $U$.

Since $N \in \mathcal F(T)$, we have $U \otimes_B T=0$ and $\operatorname{Tor}^B_1(U,T) \cong N$ as $A$-modules. Therefore, when we apply $- \otimes_B T$ to the sequence above, we obtain an exact sequence of $A$-modules $$0 \rightarrow N \rightarrow \Omega_B(U) \otimes_B T \rightarrow T' \rightarrow 0,$$ where $T'$ is in $\operatorname{add}T$.

The module $\Omega_B(U)$ is in the category $\mathcal Y(T)$ since it is a submodule of a projective $B$-module. There is an exact equivalence $- \otimes_B T \colon \mathcal Y(T) \rightarrow \mathcal T(T)$, and therefore $$\operatorname{pd}_A (\Omega_B(U) \otimes_B T) \geq \operatorname{pd}_B (\Omega_B(U))=n-1>1.$$ Since $\operatorname{pd}_A (T') \leq 1$, it follows from the exact sequence that $$\operatorname{pd}_A(N)=\operatorname{pd}_A (\Omega_B(U) \otimes_B T).$$ Combining with the previous inequality we get $$\operatorname{pd}_A(N) \geq \operatorname{pd}_B (U)-1$$ which proves the statement.