Following Qiaochu Yuan's approach, the inequalities
$$
\frac{2^n}{\log n} \ll S_n \ll \frac{2^n }{\log n}
$$
seems plausible. The lower bound is a conjecture, but it is possible to prove the upper bound.
Notations in this answer
$T_n \sim \mathrm{B}(n,\frac12)$ is the binomial distribution.
$S_n=\sum_{p\leq n} \binom np$ summed over $p$ prime.
$\pi(y)=\sum_{p\leq y}1$ is the prime counting function.
$A(n)\ll B(n)$ means $|A(n)|\leq CB(n)$ for some absolute constant $C>0$.
Lower Bound (Conjecture)
Fix $x>0$. We have
$$
P\left(p \mathrm {\ is \ prime}, \ T_n=p, \ \frac n2 -x\sqrt n\leq T_n \leq \frac n2 + x\sqrt n\right) \leq \frac {S_n}{2^n}.
$$
Since binomial coefficients $\binom nk$ peak at $k=n/2$ and become smaller when $k$ is further away from $n/2$, we take the following as a lower bound of the probability.
$$
\left(\pi(\frac n2+x\sqrt n)-\pi(\frac n2-x\sqrt n)\right)P\left(T_n=\lfloor \frac n2+x\sqrt n\rfloor\right).
$$
By Stirling's formula, and $\log (1+t)=t-\frac{t^2}2+O(\frac1{t^3})$ for $|t|\leq 1/2$, we have
$$
P\left(T_n=\lfloor \frac n2+x\sqrt n\rfloor\right)\sim \frac{2}{\sqrt{2\pi n}} e^{-2x^2}.
$$
If we have the following conjecture (see this survey by Yildrim for more information),
$$
\pi(\frac n2+x\sqrt n)-\pi(\frac n2-x\sqrt n)\sim \frac{2x\sqrt n}{\log n},
$$
then we have the conjectural lower bound
$$
\frac{4x\cdot 2^n}{e^{2x^2}\sqrt{2\pi}\log n} \lesssim S_n.
$$
Upper Bound (Easy Version)
By Hoeffding's inequality, we give a bound of sum over primes further away from $n/2$.
$$
P\left(p \mathrm {\ is \ prime}, \ T_n=p, \ |T_n-\frac n2|>\sqrt{n \log\log n} \right)
$$
$$
\leq P\left( |T_n-\frac n2|\geq \sqrt{n \log\log n}\right)\leq 2e^{-2\log\log n}\ll \frac{1}{(\log n)^2}.
$$
For the primes close to $n/2$, we apply Brun-Titchmarsh inequality,
$$
P\left(p \mathrm {\ is \ prime}, \ T_n=p, \ |T_n-\frac n2|\leq \sqrt {n \log\log n }\right) $$
$$\leq \left(\pi(\frac n2 + \sqrt {n \log\log n})-\pi(\frac n2-\sqrt {n \log\log n})\right)P\left(T_n=\lfloor \frac n2\rfloor\right)
$$
$$
\ll \frac{\sqrt{n\log\log n}}{\log n} \cdot \frac{1}{\sqrt n} = \frac{\sqrt{\log\log n}}{\log n}.
$$
Therefore, we have the upper bound
$$
S_n\ll \frac{2^n\sqrt{\log\log n}}{\log n}.
$$
Upper Bound (Added on 9/28)
With more care, we can remove $\sqrt{\log\log n}$ from the upper bound.
Again, by Hoeffding's inequality,
$$
P\left(p \mathrm {\ is \ prime}, \ T_n=p, \ |T_n-\frac n2|>\sqrt{n \log\log n} \right) \ll \frac1{(\log n)^2}.
$$
For the primes in $|T_n-\frac n2|\leq\sqrt{n \log\log n} $, consider the subintervals
$$
\frac n2 + x\sqrt n \leq p < \frac n2 + (x+1)\sqrt n
$$
for nonnegative integers $x\leq \sqrt{\log\log n}$ first.
Then the negative integers $-\sqrt{\log\log n}\leq x$ are treated similarly.
The number of primes in this interval is by Brun-Titchmarsh inequality, $\ll \frac{\sqrt n}{\log n}$, while
$$P(T_n=p)\leq P\left(T_n=\lfloor \frac n2 + x\sqrt n\rfloor\right)\sim \frac{2}{\sqrt{2\pi n}} e^{-2x^2}.$$
Note that the last asymptotic still holds if $|x|\leq \sqrt{\log\log n}$. Then we have
$$
P\left(p \mathrm {\ is \ prime}, \ T_n=p, \ \frac n2 + x\sqrt n \leq p < \frac n2 + (x+1)\sqrt n\right)
$$
$$
\ll \frac{\sqrt n}{\log n} \cdot \frac{e^{-2x^2}}{\sqrt n}.
$$
Thus by summing over $x$,
$$
P\left(p \mathrm {\ is \ prime}, \ T_n=p, \ |T_n-\frac n2|\leq \sqrt {n \log\log n }\right)$$
$$\ll \sum_{x=0}^{\infty}\frac{e^{-2x^2}}{\log n}\ll \frac 1{\log n}.
$$
Therefore, we obtain
$$
S_n\ll \frac{2^n}{\log n}.
$$
Update on 2019/3/4
Nilotpal Kanti Sinha and I started working on writing a paper on this subject. Here is current progress. The proofs are too long to be contained here, but the main idea of splitting up the sum into short intervals is present in this answer. To prove 1, we need Huxley's zero density estimate and its consequence on the primes in the short intervals. (Chapter 5 of this note by Angel Kumchev: https://tigerweb.towson.edu/akumchev/a5.pdf).
- We have for almost all $n$,
$$
S_n= \frac{2^n}{\log n}+O\left(\frac{2^n}{(\log n)^2}\right) \ \textrm{as }n\rightarrow \infty.
$$
Here, almost all means that the number of $n\in [1,N]\cap \mathbb{Z}$ for which the asymptotic formula fails is $o(N)$.
We have
$$
\alpha:=\liminf_{n\rightarrow\infty}\frac{S_n\log n}{2^n}\leq 1\leq \limsup_{n\rightarrow\infty} \frac{S_n \log n}{2^n} \leq 4.
$$
The statement $\alpha>0$ implies that, there is $b>0$ and $N_0(b)>0$ such that,
$$
\pi\left(\frac n2 +\sqrt {n\log\log n}\right)-\pi\left( \frac n2-\sqrt {n\log\log n}\right)\geq \frac{b\sqrt n}{\log n} \ \textrm{for all }n\geq N_0(b).
$$
Best Answer
This is an interesting problem from a numerical point of view.
Transposed in the real domain, you are looking for $m$ such that
$$\color{blue}{\frac{\, _2F_1(1,m-n+1;m+2;-1)}{\Gamma (m+2) \,\,\Gamma (n-m)} =\frac{2^n}{\Gamma (n+1)}-\frac{1}{\Big[\Gamma \left(\frac{n}{2}+1\right)\Big]^2}}$$ where $m$ and $n$ are real numbers.
This equation is not difficult to solve using Newton method with $m_0=\frac n 2$. This starting point is justified by the left part of the trivial double inequality $$\binom{n}{m} \leq\sum_{k=0}^m\binom{n}{k}\leq (m+1)\binom{n}{m}$$ which means that we already know that $m \leq \frac n 2$. I did not find any simple way to use the right part of the above inequality (this is no more true : have a look at the $\color{red}{\text{ update}}$) .
For example, for $n=10$, the iterates are $$\left( \begin{array}{cc} k & m_k \\ 0 & 5.000000000 \\ 1 & 3.419647982 \\ 2 & 3.407971414 \\ 3 & 3.407943361 \end{array} \right)$$
Below are some results (I let you rounding the results the way you want). $$\left( \begin{array}{cc} n & m \\ 10 & 3.40794 \\ 20 & 7.41879 \\ 30 & 11.5964 \\ 40 & 15.8702 \\ 50 & 20.2093 \\ 60 & 24.5969 \\ 70 & 29.0227 \\ 80 & 33.4793 \\ 90 & 37.9619 \\ 100 & 42.4665 \\ 110 & 46.9903 \\ 120 & 51.5309 \\ 130 & 56.0864 \\ 140 & 60.6554 \\ 150 & 65.2365 \\ 160 & 69.8287 \\ 170 & 74.4310 \\ 180 & 79.0426 \\ 190 & 83.6628 \\ 200 & 88.2910 \\ 210 & 92.9267 \\ 220 & 97.5694 \\ 230 & 102.219 \\ 240 & 106.874 \\ 250 & 111.535 \\ 260 & 116.202 \\ 270 & 120.874 \\ 280 & 125.550 \\ 290 & 130.232 \\ 300 & 134.918 \end{array} \right)$$
This looks to be very close to linearity. Using these numbers, a quick and dirty linear regression for $m=a +b \,n$ leads to $R^2=0.999957$ $$\begin{array}{clclclclc} \text{} & \text{Estimate} & \text{Standard Error} & \text{Confidence Interval} \\ a & -2.6778 & 0.2028 & \{-3.0938,-2.2618\} \\ b & +0.4561 & 0.0011 & \{+0.4538,+0.4585\} \\ \end{array}$$
Using this empirical model for $n=400$, it gives $m=179.775$ while the solution is $181.986$.
As I wrote in comments, this also works for non integer values of $n$. For $n=123.456$, $m=53.1037$.
Update
I managed to use $$\sum_{k=0}^m\binom{n}{k}\leq (m+1)\binom{n}{m}$$ defining the function $$f(m)=(m+1)\binom{n}{m}- {n\choose \frac{n}{2}}$$ which was expanded as a series to $O\left(\left(m-\frac{n}{2}\right)^3\right)$. Solving the quadratic, the approximate solution is given by $$m=\frac n 2-\frac{n}{1+\sqrt{n} \sqrt{(n+2) \psi ^{(1)}\left(\frac{n}{2}\right)-\frac{3 n+8}{n^2}}}$$ which is a much better starting point as shown below $$\left( \begin{array}{ccc} n & \text{approximation} & \text{solution} \\ 50 & 20.5142 & 20.2093 \\ 100 & 43.4413 & 42.4665 \\ 150 & 66.8507 & 65.2365 \\ 200 & 90.5099 & 88.2910 \\ 250 & 114.329 & 111.535 \\ 300 & 138.261 & 134.918 \end{array} \right)$$
The asymptotics of the approximation is $$m=\frac n2 \left(1-\sqrt{\frac 2 n}+\frac 1 n+O\left(\frac{1}{n^{3/2}}\right)\right)$$
Update
I found later this question; @user940 gave a very interesting asymptotic approximation. Adapted to your problem, we look for the solution $m$ of the equation $$2^{n-1} \left(1-\text{erf}\left(\frac{n-2 m}{\sqrt{2n} }\right)\right)=\binom{n}{\frac{n}{2}}$$ that is to say $$\text{erf}\left(\frac{n-2 m}{\sqrt{2n} }\right)=1-\frac{2\, \Gamma \left(\frac{n+1}{2}\right)}{\sqrt{\pi } \,\Gamma \left(\frac{n}{2}+1\right)}$$ This can be inversed using approximations of the error function (have a look here).
This would give $$\left( \begin{array}{cc} 50 & 20.7060 \\ 100 & 42.9608 \\ 150 & 65.7299 \\ 200 & 88.7840 \\ 250 & 112.028 \\ 300 & 135.410 \end{array} \right)$$ which is significantly better for large values of $n$.
Concerning the asymptotics of $n$, using $$\text{erf}(x)=1+e^{-x^2} \left(-\frac{1}{\sqrt{\pi } x}+O\left(\frac{1}{x^3}\right)\right)$$ we have $$m=\frac n 2-\frac {\sqrt n } 2 \sqrt{W(t)}\qquad \text{where} \qquad t=\frac 12\left(\frac{\Gamma \left(\frac{n+2}{2}\right)}{\Gamma \left(\frac{n+1}{2}\right)} \right)^2$$ $W(.)$ being Lambert function. So, as expected earlier, a logarithmic contribution in the asymptotics of $m$.