= Solution
There is no minimum about which this scalar can undergo the damped oscillations normally used for <reheating>. Its slow-roll parameters are constant, so an inflating exponential solution has no automatic exit. An independent stopping and energy-transfer mechanism is needed.
At the assumed exit, $V_R=m_{\rm pl}^4e^{-30}$. Equating this to the thermal radiation <density> gives
$$
\boxed{T_R\simeq\left(\frac{30}{\pi^2g_*}\right)^{1/4}m_{\rm pl}e^{-30/4}\simeq1.5\times10^{15}\ {\rm GeV}}
$$
for $g_*=10^3$. Corrections from the small kinetic fraction do not change this rough estimate.
<Curvature> obeys $|\Omega-1|\propto(aH)^{-2}$. For constant $\epsilon_H=A^2/(16\pi)$ and $N=\log(a_R/a_i)$, its suppression during <cosmic inflation> is $e^{-2(1-\epsilon_H)N}$. Starting from order-one <curvature> and requiring present <curvature> no larger than order one gives
$$
N\gtrsim\frac{\log(a_RH_R/a_0H_0)}{1-\epsilon_H}.
$$
After instantaneous <reheating>, entropy conservation implies $a_R/a_0=(T_0/T_R)(g_{*s,0}/g_{*s,R})^{1/3}$, with $H_R\simeq1.66\sqrt{g_*}T_R^2/m_{\rm pl}$. Present scales make the numerator roughly $60$; using the literal approximate units from question2 gives about $63$. A tighter present <curvature> tolerance adds its corresponding half-logarithm. In <slow-roll approximation> this is the conventional requirement of roughly sixty e-folds.
Because the field increases toward $\phi_R$, enough expansion requires
$$
\boxed{\phi_i\lesssim\frac{30m_{\rm pl}}A-\frac{Am_{\rm pl}}{8\pi}\frac{60}{1-A^2/(16\pi)}}.
$$
This is the <flatness bound on the initial field in exponential inflation>. It is an \b[upper bound on the initial field], not a nonzero minimum: starting earlier at a smaller field produces more expansion. If one additionally restricts the initial potential below $m_{\rm pl}^4$, then $\phi_i\ge0$ supplies a lower bound, but that is a separate restriction. Without initial-curvature and final-tolerance choices there is no unique numerical initial-field value.
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