Define $s(J)=\sum_{j\in J}j$ and $[n]=\{1,...,n\}$. I'm trying to get some reasonable upper bound on $$\sum_{J\subset [n], |J|=i}\frac{1}{2^{s(J)}}.$$
Actually, I want an upper bound on $$\sum_{i=k}^{2k}\sum_{J\subset [n], |J|=i}\frac{1}{2^{s(J)}}$$ for a fixed $k<\frac{n}{2}$.
The best I could get was $$\sum_{i=k}^{2k}\sum_{J\subset [n], |J|=i}\frac{1}{2^{s(J)}}\leq \sum_{i=0}^{n}\sum_{J\subset [n], |J|=i}\frac{1}{2^{s(J)}}=\prod_{i=1}^n \left(1+\frac{1}{2^i}\right).$$
I'd appreciate if someone could find anything better than that, preferentially depending on $k$.
Edit: Using $$\sum_{J\subset [n], |J|=i}\frac{1}{2^{s(J)}}\leq \sum_{J\subset [n], |J|=i}\frac{1}{2^{\frac{i(i+1)}{2}}}$$ wasn't enough for me either.
Here is another crude bound: Your sum
$$ S(n,j) := \sum_{\substack{J \subseteq [n] \\ |J| = j}} \prod_{k \in J} \frac{1}{2^k} $$
can be bounded from above by
$$ S(n, j) \leq \frac{1}{2^{j(j+1)/2}} \prod_{k=1}^{j} \frac{1}{1 - 2^{-k}}. \tag{*} $$
To derive this, we may write
\begin{align*} S(n,j) &= \sum_{0<k_1<\dots<k_j\leq n} \frac{1}{2^{k_1 + \dots + k_j}} \\ &= \frac{1}{2^{j(j+1)/2}} \sum_{0 \leq r_1 \leq \dots \leq r_j \leq n-j} \frac{1}{2^{r_1 + \dots + r_j}} \tag{$k_i = i + r_i$} \\ &= \frac{1}{2^{j(j+1)/2}} \sum_{\lambda} \frac{1}{2^{|\lambda|}}, \end{align*}
where the sum runs over all partitions $\lambda = (\lambda_1, \dots, \lambda_l)$ for which the length $l$ is at most $n-j$ and each part $\lambda_i$ is at most $j$. Also, $|\lambda| = \lambda_1 + \dots + \lambda_l $ denotes the size of $\lambda$. Then by relaxing the restriction on the length $l$, we obtain the desired bound $\text{(*)}$.
This bound seems useful if $j \ll n$. Indeed, a loose bound
$$ \left| \prod_{k=1}^{j} \frac{1}{1 - 2^{-k}} - \sum_{\lambda} \frac{1}{2^{|\lambda|}} \right| \leq \frac{C n^j}{2^{n-j} - 1} \tag{2} $$
for the constant $C = \prod_{k=1}^{\infty} (1 - 2^{-k})^{-1} \approx 3.46275$ is already enough to prove that the relative error between $S(n,j)$ and the upper bound in $\text{(*)}$ decays exponentially fast along the limit as $n\to\infty$ and $j=\mathcal{O}(\log n)$.
Addendum. Proof of $\text{(2)}$: We have
$$ \prod_{k=1}^{j} \frac{1}{1 - 2^{-k}} - \sum_{\lambda} \frac{1}{2^{|\lambda|}} = \sum_{\mu} \frac{1}{2^{|\mu|}}, $$
where the sum in the right-hand side runs over all partitions $\mu$ for which the length is $>n-j$ and each part is $\leq j$. Now for each $l = 1, \dots, j$, the contribution from all such partitions $\mu$ with $\mu_{n-j} = l$ is
\begin{align*} \sum_{\mu \,:\, \mu_{n-j} = l} \frac{1}{2^{|\mu|}} &= \sum_{\substack{j \geq \mu_1 \geq \mu_2 \geq \dots \\ \mu_{n-j} = l}} \frac{1}{2^{\mu_1 + \mu_2 + \dots}} \\ &= \Biggl( \sum_{j \geq \mu_1 \geq \dots \geq \mu_{n-j-1} \geq l} \frac{1}{2^{\mu_1 + \dots + \mu_{n-j-1} + l}} \Biggr) \Biggl(\sum_{l \geq \mu_{n-j+1} \geq \dots} \frac{1}{2^{\mu_{n-j+1} + \dots}}\Biggr). \end{align*}
Now the first sum in the last step can be bounded from above by
$$ \frac{\#\{ (\mu_1,\dots,\mu_{n-j+1}) : j \geq \mu_1 \geq \dots \geq \mu_{n-j-1} \geq l \}}{2^{(n-j)l}} \leq \frac{n^j}{2^{(n-j)l}}, $$
and the second sum satisfies
$$ \sum_{l \geq \mu_{n-j+1} \geq \dots} \frac{1}{2^{\mu_{n-j+1} + \dots}} = \prod_{k=1}^{l} \frac{1}{1 - 2^{-k}} \leq C. $$
Therefore
$$ \sum_{\mu} \frac{1}{2^{|\mu|}} \leq \sum_{l=1}^{j} \frac{Cn^j}{2^{(n-j)l}} \leq \sum_{l=1}^{\infty} \frac{Cn^j}{2^{(n-j)l}} = \frac{Cn^j}{2^{n-j}-1}. $$