It is known that a MOS transistor with the source-body voltage different from zero has the threshold voltage modified by the body effect, that is if $ V_{sb}\neq 0 $, with $ V_{sb}$ the source-body voltage (let's remember that for a n-MOSFET $ V_b = V_{SS}$ and for a p-MOSFET $ V_b = V_{DD}$), then $ \vert V_{Th}\vert _{V_{sb} \neq 0} > \vert V_{Th}\vert _{V_{sb} = 0}$. The initial conditions of the chain nodes are set by the initial condition on the output. So if the output node is discharging, then one (and only one) n-MOSFET is switching from off to on. It means that all the other MOSFETs are already on, and while the starting voltage of the output node is $ V_{DD}$, all the internal nodes have as a starting voltage $ V_{DD} - V_{Tn}^{*}$.
With the notations of previous paragraphs, the N-th (topmost) n-MOS transistor has $ V_{s_n}^N = V_{DD} - V_{Tn}^{*}$, with $ V_s$ source potential and $ V_{Tn}^*$ the threshold voltage modified by the body effect. All the internal transistors have $ V_{d_n}^i=V_{s_n}^i=V_{DD}-V_{Tn}^* $, while the first one has $ V_{d_n}^1=V_{DD}-V_{Tn}^*$ and $ V_{s_n}^1=0$.
The threshold voltage variation as a function of $ V_{sb}$ is given by:
$\displaystyle \Delta V_{Tn}=\gamma(\sqrt{2\vert\Phi_p\vert+V_{sb}} - \sqrt{2\vert\Phi_p\vert})\; ,$
with $ \gamma= \frac{\sqrt{2\varepsilon_s q N_a}}{C_{ox}}$ and $ \Phi_p = -\frac{KT}{q}\ln{(\frac{N_a}{n_i})}.$
The source potential of the top transistor is
$\displaystyle V_s = V_{DD}-V_{Tn}^*\; ,$
and, if $ V_{Tn0}$ is the threshold voltage with $ V_{sb}=0$, then $ V_{Tn}^*=V_{Tn0}+\Delta V_{Tn}$ and we can solve for $ V_{sb}$:
\begin{displaymath}\begin{split}V_{sb}&= \pm \frac{\gamma\sqrt{4\gamma\sqrt{2\ve... ... + V_{DD} - V_{Tn0} + \frac{\gamma^2}{2} \quad (>0) \end{split}\end{displaymath}