I will only give some simple and useful estimate (confirmed by numerical results, based on some mathematical idea). It is essentially a lower bound, but can probably be easily modified to be an upper bound. Moreover, it might give you further ideas.
Denote your expression by $\varphi_n$, and consider a random variable $X$ with binomial$(n,1/2)$ distribution. Then
$$
{\rm E}\big[2^{\sqrt X } \big] = \sum\limits_{k = 0}^n {2^{\sqrt k } {n \choose k}\frac{1}{{2^n }}} = \frac{1}{{2^n }}\varphi _n.
$$
The left-hand side is actually the moment-generating function $M(t)$ of $\sqrt{X}$ at $t=\ln2$, but we will not use this fact.
I have plotted the function $2^{\sqrt{x}}$, $x > 0$, and it seems that this function is typically convex (it is not convex near $0$). So, heuristically, Jensen's inequality suggests the following result:
$$
{\rm E}\big[2^{\sqrt X } \big] > 2^{\sqrt {{\rm E}[X]} },
$$
for all sufficiently large $n$. (If $n$ is large, then $X$ typically takes values in a set on which $2^{\sqrt{x}}$ is convex.) Finally, ${\rm E}[X]=n/2$ leads to $2^{ - n} \varphi _n > 2^{\sqrt {n/2} }$, that is
$$
\sum\limits_{k = 0}^n {{n \choose k} 2^{\sqrt k } } > 2^{n + \sqrt {n/2} },
$$
for all sufficiently large $n$. Numerical results indicate that this inequality holds for all $n \geq 6$. Moreover, the results suggest that the (lower-)bound is quite tight (I checked for $n \leq 150$). So, a tight upper-bound is likely to be something not far from $2^{n + \sqrt {n/2}}$.
EDIT: We can also employ the strong law of large numbers (SLLN) to show that $\varphi_n$ may be expected to behave something like $2^{n + \sqrt {n/2}}$ for large $n$. Indeed, writing $X$ as $Y_1 + \cdots + Y_n$, where $Y_i$ are i.i.d. rv's with ${\rm P}(Y_i = 0) = {\rm P}(Y_i = 1) = 1/2$, we have
$$
\varphi _n = 2^n {\rm E}\bigg[2^{\sqrt n \sqrt {\frac{{Y_1 + \cdots + Y_n }}{n}} } \bigg].
$$
By SLLN, $\frac{{\sum\nolimits_{i = 1}^n {Y_i } }}{n}$ converges to $1/2 \,(= {\rm E}[Y_1])$ almost surely, and the conclusion follows.
The Lagrange Inversion Formula provides an appropriate method to derive
\begin{align*}
\sum_{k=0}^\infty \binom{3k}{k}\frac{1}{8^k}=\frac{4\sqrt{10}}{5}\cos\Big(\frac{1}{3}\arcsin \frac{3\sqrt{6}}{8}\Big)\tag{1}
\end{align*}.
Let a formal power series $w=w(t)$ be implicitely defined by a relation $w=t\Phi(w)$, where $\Phi(t)$ is a formal power series such that $\Phi(0)\ne0$. The Lagrange Inversion Formula (LIF) states that:
$$[t^n]w(t)^k=\frac{k}{n}[t^{n-k}]\Phi(t)^n$$
A variation stated as formula $G6$ in Lagrange Inversion: when and how by R. Sprugnoli (etal) is:
Let $F(t)$ be any formal power series and $w=t\Phi(w)$ as before, then the following is valid:
\begin{align*}
[t^n]F(t)\Phi(t)^n=\left[\left.\frac{F(w)}{1-t\Phi'(w)}\right|w=t\Phi(w)\right]\tag{2}
\end{align*}
Note: The notation $[\left.f(w)\right|w=g(t)]$ is a linearization of $\left.f(w)\right|_{w=g(t)}$ and denotes the substitution of $g(t)$ to every occurrence of $w$ in $f(w)$ (that is, $f(g(t))$). In particular, $w=t\Phi(w)$ is to be solved in $w=w(t)$ and $w$ has to be substituted in the expression on the left of the $|$ sign.
In order to prove (1) we set $F(t)=1$ and $\Phi(t)=(1+t)^3$. We then have
$$t\Phi'(w)=3t(1+w)^2=\frac{3t\Phi(w)}{1+w}=\frac{3w}{1+w}$$
It follows:
\begin{align*}
\binom{3n}{n}&=[t^n]F(t)\Phi(t)^n=[t^n](1+t)^{3n}\\
&=[t^n]\left[\left.\frac{1}{1-t\Phi'(w)}\right|w=t\Phi(w)\right]\\
&=[t^n]\left[\left.\frac{1}{1-\frac{3w}{1+w}}\right|w=t\Phi(w)\right]\\
&=[t^n]\left[\left.\frac{1+w}{1-2w}\right|w=t\Phi(w)\right]\\
\end{align*}
Let
\begin{align*}
A(t):=\sum_{n\ge0}\binom{3n}{n}t^n=\left.\frac{1+w}{1-2w}\right|_{w=t\Phi(w)}
\end{align*}
Expressing $A(t)=\frac{1+w}{1-2w}$ in terms of $w$, we get
$$w=\frac{A(t)-1}{2A(t)+1}$$
Since $w=t\Phi(w)=t(1+w)^3$, we obtain
\begin{align*}
\frac{A(t)-1}{2A(t)+1}=t\left(1+\frac{A(t)-1}{2A(t)+1}\right)^3
\end{align*}
which simplifies to:
\begin{align*}
(4-27t)A(t)^3-3A(t)-1=0\tag{3}
\end{align*}
In order to get the RHS of $(1)$ we first analyse the structure of (3) which is
$$f(t)A(t)^3-3A(t)=1$$
with $f(t)$ linear and observe a similarity of this structure with the identity
$$4\cos^3{t}-3\cos{t}=\cos{3t}$$
Thus we use the ansatz:
\begin{align*}
A(t) := \frac{2\cos\left(g(t)\right)}{\sqrt{4-27t}}\tag{4}
\end{align*}
We see
\begin{align*}
(4-27t)A(t)^3-3A(t)
&=\frac{8\cos^3\left(g(t)\right)}{\sqrt{4-27t}}-\frac{6\cos\left(g(t)\right)}{\sqrt{4-27t}}=\\
&=\frac{2\cos\left(3g(t)\right)}{\sqrt{4-27t}}\\
&=1
\end{align*}
Now, since
\begin{align*}
2\cos\left(3g(t)\right)&=\sqrt{4-27t}\\
4\cos^2\left(3g(t)\right)&=4-27t\\
\sin^2\left(3g(t)\right)&=\frac{27}{4}t\\
\end{align*}
we get
\begin{align*}
g(t)&=\frac{1}{3}\arcsin\left(\frac{3\sqrt{3t}}{2}\right)\tag{5}\\
\end{align*}
We finally conclude from (4) and (5)
\begin{align*}
\color{blue}{\sum_{k=0}^\infty \binom{3k}{k}\frac{1}{8^k}}
&=\left.\frac{2\cos(g(t))}{\sqrt{4-27t}}\right|_{t=\frac{1}{8}}\\
&=\left.\frac{2\cos\left(\frac{1}{3}\arcsin\left(\frac{3\sqrt{3t}}{2}\right)\right)}{\sqrt{4-27t}}\right|_{t=\frac{1}{8}}\\
&\,\,\color{blue}{=\frac{4\sqrt{10}}{5}\cos\left(\frac{1}{3}\arcsin\left( \frac{3\sqrt{6}}{8}\right)\right)}
\end{align*}
and the claim follows.
Best Answer
By (one of the formulations of) Stirling approximation formula, you have $n!=(2\pi n)^{1/2}(n/e)^{n}e^{\theta(n)}$, with $\frac{1}{12n+1}\le \theta(n) \le \frac{1}{12n}$. Therefore, you can compute
$$(2n)!=2^{2n+1}(\pi n)^{1/2}(n/e)^{2n}e^{\theta(2n)}=2^{2n}\frac{n!^2}{(\pi n)^{1/2}} e^{\theta(2n)-2\theta(n)},$$
that is
$$ \binom{2n}{n}= \frac{2^{2n}}{(\pi n)^{1/2}} e^{\theta(2n)-2\theta(n)}.$$
Now, noting that
$\theta(2n)-2\theta(n)\le\frac{1}{24n}-\frac{2}{12n+1}<0$
and
$\theta(2n)-2\theta(n)\ge \frac{1}{24n+1}-\frac{2}{12n}\ge -\frac{1}{6}>-\log 2$,
you can conclude.