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.