The first eigenvalue $\lambda_1$ is of order $T^{-2}$. One can get the upper bound by a direct analysis of the Rayleigh quotient. Let $M_\pm$ be the two ends. First suppose $\operatorname{Vol}(M_+)=\operatorname{Vol}(M_-)$; then the function
$$
f(x)=\begin{cases}\sin(\tfrac{1}{2}\pi t/T),& \text{for $x=(s,t)$ in the neck part $S\times [-T,T]$}\\\pm 1,& x\in M_\pm
\end{cases}
$$
is in the function space $H^1(M)$ and $\int_Mf=0$, so
$$
\lambda_1\leq \frac{\int_M|df|^2}{\int_Mf^2}
=\frac{(\tfrac{\pi}{2T})^2T\operatorname{Vol}(S)}{T\operatorname{Vol}(S)+2\operatorname{Vol}(M_\pm)}\sim\frac{\pi^2}{4T^2}.
$$
If instead $\operatorname{Vol}(M_+)>\operatorname{Vol}(M_-)$ then let $c:=[\operatorname{Vol}(M_+)-\operatorname{Vol}(M_-)]/\operatorname{Vol}(S)$, add the neck portion $S\times [-T, -T+c]$ to $M_-$ to equalize the volumes, and run the above argument with $T-\tfrac{1}{2}c$ replacing $T$.

*Remark:** Edited to add the above argument, which improves by an asymptotic factor of 4 the upper bound in my original answer.*

For the lower bound, as noted by @Neal, it seems that the Cheeger constant $h(M)$ should be achieved by a cross-section $S$ in the neck part, so
$$h(M)=\frac{\operatorname{Vol}(S)}{\operatorname{Vol}(M)/2}
=\frac{\operatorname{Vol}(S)}{T\operatorname{Vol}(S)+c}\sim \frac{1}{T}.
$$
This means that $\lambda_1\geq h(M)^2/4\sim\tfrac{1}{4}T^{-2}$.