The original paper addresses this issue on p. 83:
$$
p(n)=\frac{1}{2\pi\sqrt2}\frac{d}{dn}\left(\frac{e^{C\lambda_n}}{\lambda_n}\right) + \frac{(-1)^n}{2\pi}\frac{d}{dn}\left(\frac{e^{C\lambda_n/2}}{\lambda_n}\right) + O\left(e^{(C/3+\varepsilon)\sqrt n}\right)
$$
with
$$
C=\frac{2\pi}{\sqrt6},\ \lambda_n=\sqrt{n-1/24},\ \varepsilon>0.
$$
If I compute correctly, this gives
$$
e^{\pi\sqrt{\frac{2n}{3}}} \left(
\frac{1}{4n\sqrt3}
-\frac{72+\pi^2}{288\pi n\sqrt{2n}}
+\frac{432+\pi^2}{27648n^2\sqrt3}
+O\left(\frac{1}{n^2\sqrt n}\right)
\right)
$$
The argument you give generalizes to show that
$$p(n) \ge (k+1)^{ \lfloor \sqrt{ n/k } \rfloor }$$
for any positive integer $k$. We repeat the same argument, but instead of subsets of $\{ 1, 2, ... \lfloor \sqrt{n} \rfloor \}$ we allow multisubsets of $\{ 1, 2, ... \lfloor \sqrt{n/k} \rfloor \}$, where each element occurs with multiplicity $0$ through $k$.
So how much better is this? The actual asymptotic value of $\log p(n)$ is known to be $\pi \sqrt{ 2n/3 }$, so $C \sqrt{n}$ where $C \approx 2.565...$. The argument you give shows that $C \ge \log 2 \approx 0.693...$. The generalization above shows that $C \ge \frac{\log (k+1)}{\sqrt{k}}$. Some numerical experimentation shows that this attains a maximum at $k = 4$, giving $C \ge \frac{\log 5}{2} \approx 0.805...$. So this is a little better, but still a long way to go.
Edit #2: Here is an argument along the above lines which gives $C \ge \sqrt{2} \approx 1.414...$. We will now allow the positive integer $k$ to occur with a multiplicity from $0$ to $m_k - 1$ for some $m_k \ge 1$. This produces a partition in the same way as above provided that the sum constraint $\sum k (m_k - 1) \le n$ is satisfied, and we get $\prod m_k$ partitions this way. A heuristic calculation using Lagrange multipliers suggests that we want $m_k$ to be proportional to $\frac{1}{k}$; we will in fact take
$$m_k = \left\lfloor \frac{m}{k} \right\rfloor$$
if $k \le t$ (and $m_k = 1$ otherwise) for some $m, t$ satisfying certain constraints. First, since we must have $m_k \ge 1$, we need $m \ge t$. Second, the sum constraint $\sum k (m_k - 1) \le n$ gives
$$mt - \frac{t(t+1)}{2} \le n.$$
We will choose $m, t$ later. First, taking the logarithm of the number of partitions gives
$$\sum_{k=1}^t \log \left\lfloor \frac{m}{k} \right\rfloor.$$
We can write this as approximately (ignoring floors now)
$$\sum_{k=1}^t (\log m - \log k) = t \log m - \sum_{k=1}^t \log k.$$
(This is an overestimate but I believe it does not affect the asymptotic.) The estimate $\log n! \approx n \log n - n + O(\log n)$ gives that this is asymptotically
$$t \left( \log \frac{m}{t} + 1 \right) + O(\log t).$$
Now to optimize $m, t$. First, we can replace the sum constraint with a constraint
$$mt - \frac{t^2}{2} \le n$$
which is easier to deal with. Taking $t = a \sqrt{n}$ gives
$$m \le \left( \frac{1}{a} + \frac{a}{2} \right) \sqrt{n}$$
so we can just take $m$ to be this value (ignoring floors again); the constraint $m \ge t$ is now equivalent to the constraint $a \le \sqrt{2}$. The logarithm of the number of partitions we get now takes the form
$$a \sqrt{n} \left( \log \left( \frac{1}{2} + \frac{1}{a^2} \right) + 1 \right) + O(\log n)$$
which gives
$$C \ge a \left( \log \left( \frac{1}{2} + \frac{1}{a^2} \right) + 1 \right).$$
Some numerical experiments suggest that the RHS is increasing for $a \le \sqrt{2}$, and taking $a = \sqrt{2}$ gives $C \ge \sqrt{2}$ as desired.
I do not think this type of argument can be pushed substantially further without some new idea.
Best Answer
Using the leading term of Radmacher's formula, we have \begin{align*} p(n) & \sim \frac{1}{{\pi \sqrt 2 }}\left[ {\frac{\mathrm{d}}{{\mathrm{d}x}}\frac{{\sinh \left( {\pi \sqrt {\frac{2}{3}\left( {x - \frac{1}{{24}}} \right)} } \right)}}{{\sqrt {x - \frac{1}{{24}}} }}} \right]_{x = n} \\ & = \frac{{4\sqrt 3 }}{{24n - 1}}\cosh \left( {\pi \sqrt {\frac{2}{3}\left( {n - \frac{1}{{24}}} \right)} } \right) - \frac{1}{\pi}\frac{{24\sqrt 3 }}{{(24n - 1)^{3/2} }}\sinh \left( {\pi \sqrt {\frac{2}{3}\left( {n - \frac{1}{{24}}} \right)} } \right). \end{align*} Now \begin{align*} \pi \sqrt {\frac{2}{3}\left( {n - \frac{1}{{24}}} \right)} = \pi \sqrt {\frac{2}{3}n} \sqrt {1 - \frac{1}{{24n}}} & = \pi \sqrt {\frac{2}{3}n} \left( {1 + \mathcal{O}\!\left( {\frac{1}{n}} \right)} \right) \\ &= \pi \sqrt {\frac{2}{3}n} + \mathcal{O}\!\left( {\frac{1}{{\sqrt n }}} \right). \end{align*} Thus, \begin{align*} & \cosh \left( {\pi \sqrt {\frac{2}{3}\left( {n - \frac{1}{{24}}} \right)} } \right),\sinh \left( {\pi \sqrt {\frac{2}{3}\left( {n - \frac{1}{{24}}} \right)} } \right) \sim \frac{1}{2}\exp \left( {\pi \sqrt {\frac{2}{3}\left( {n - \frac{1}{{24}}} \right)} } \right) \\ & = \frac{1}{2}\exp \left( {\pi \sqrt {\frac{2}{3}n} + \mathcal{O}\!\left( {\frac{1}{{\sqrt n }}} \right)} \right) = \frac{1}{2}\exp \left( {\pi \sqrt {\frac{2}{3}n} } \right)\left( {1 +\mathcal{O}\!\left( {\frac{1}{{\sqrt n }}} \right)} \right) \\ & \sim \frac{1}{2}\exp \left( {\pi \sqrt {\frac{2}{3}n} } \right). \end{align*} And therefore, \begin{align*} p(n) & \sim \frac{{4\sqrt 3 }}{{24n - 1}}\frac{1}{2}\exp \left( {\pi \sqrt {\frac{2}{3}n} } \right) - \frac{1}{\pi }\frac{{24\sqrt 3 }}{{(24n - 1)^{3/2} }}\frac{1}{2}\exp \left( {\pi \sqrt {\frac{2}{3}n} } \right) \\ & \sim \frac{{4\sqrt 3 }}{{24n}}\frac{1}{2}\exp \left( {\pi \sqrt {\frac{2}{3}n} } \right) - \frac{1}{\pi }\frac{{24\sqrt 3 }}{{(24n)^{3/2} }}\frac{1}{2}\exp \left( {\pi \sqrt {\frac{2}{3}n} } \right) \\ & = \frac{1}{{4n\sqrt 3 }}\exp \left( {\pi \sqrt {\frac{2}{3}n} } \right)\left( {1 - \frac{{\sqrt 3 }}{{\sqrt 2 \pi n^{1/2} }}} \right) \sim \frac{1}{{4n\sqrt 3 }}\exp \left( {\pi \sqrt {\frac{2}{3}n} } \right). \end{align*}