Under your conditions, $P(|r|\leq\frac12\big(1+\frac1{n^c}\big)b^2T)$ will usually go to $0$ as $n\to\infty$. So, the lower bound $1-\frac1{n^2}$ on this probability will be impossible.
Indeed, for simplicity let $T=1$. Note that $c_b^2:=Var(v_i)=(b^2-1)/12$. Let $u_{ij}:=t_i t_j$, where $t_1,\dots,t_n$ are iid Rademacher random variables (independent of the $v_i$'s), with $P(t_i=\pm1)=1/2$. Let $n\to\infty$. Then \begin{equation} \frac rn=\Big(\frac1{\sqrt n}\sum_{i=1}^n t_i v_i\Big)^2\to c_b^2 Z^2 \end{equation} in distribution, by the central limit theorem, where $Z\sim N(0,1)$. So, for any real $\delta>0$ and all $n>b^2/\delta$, \begin{equation} P(|r|\leq\tfrac12\big(1+\tfrac1{n^c}\big)b^2T) \le P(\tfrac{|r|}n\le\tfrac{b^2}n) \le P(\tfrac{|r|}n\le\delta)\to P(c_b^2 Z^2\le\delta), \end{equation} and the latter probability goes to $0$ as $\delta\downarrow0$.