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Define the complex-variable function $$g(z)=\frac{\sin\pi z}{\pi z}\prod_{k=1}^n\frac1{1-\frac{z}k}.$$ Note $f(k)=\binom{n}k$ for $k\in\mathbb{Z}^{\geq0}$. Moreover, $\sqrt{g(z)}$ is analytic in the region $-1<\Re(z)<n+1$ since $g(z)$ is zero-free there. By the [Residue Theorem][1], $$\sum_{k=0}^n\sqrt{\binom{n}k} =\frac1{2\pi i}\int_C\sqrt{g(z)}\,\frac{2\pi}{\sin2\pi z}\,dz$$ where $C$ is a closed (positive) contour inside $-1<\Re(z)<n+1$ going around once about each integer $0\leq k\leq n$. Choose $C_R$ as the rectangular along the lines $\Re(z)=-\frac12$, $\Re(z)=n+\frac12$ and $\Im(z)=\pm R$. Thus $$\int_{C_R}\sqrt{g(z)}\,\frac{2\pi}{\sin2\pi z}\,dz =\int_{C_R}\prod_{k=1}^n\sqrt{\frac1{1-\frac{z}k}}\,\frac{\sqrt{\pi}\,dz} {\cos\pi z\,\sqrt{z\sin\pi z}}.$$ So, letting $R\rightarrow\infty$, we arrive at $$\sum_{k=0}^n\sqrt{\binom{n}k} =\frac1{2\pi i}\left[\int_{-\frac12+i\infty}^{-\frac12-i\infty} +\int_{n+\frac12-i\infty}^{n+\frac12+i\infty} \sqrt{g(z)}\,\frac{2\pi}{\sin2\pi z}\,dz\right].$$ The integrand is invariant under $z\rightarrow n-z$, hence it suffices to compute (actually estimate) one of the integrals.

This is where I left off. Perhaps someone can complete the argument. [1]: https://en.wikipedia.org/wiki/Residue_theorem

Define the complex-variable function $$g(z)=\frac{\sin\pi z}{\pi z}\prod_{k=1}^n\frac1{1-\frac{z}k}.$$ Note $f(k)=\binom{n}k$ for $k\in\mathbb{Z}^{\geq0}$. Moreover, $\sqrt{g(z)}$ is analytic in the region $-1<\Re(z)<n+1$ since $g(z)$ is zero-free there. By the [Residue Theorem][1], $$\sum_{k=0}^n\sqrt{\binom{n}k} =\frac1{2\pi i}\int_C\sqrt{g(z)}\,\frac{\pi\cos\pi z}{\sin\pi z}\,dz$$ where $C$ is a closed (positive) contour inside $-1<\Re(z)<n+1$ going around once about each integer $0\leq k\leq n$. Choose $C_R$ as the rectangular along the lines $\Re(z)=-\frac12$, $\Re(z)=n+\frac12$ and $\Im(z)=\pm R$. Thus $$\int_{C_R}\sqrt{g(z)}\,\frac{\pi\cos\pi z}{\sin\pi z}\,dz =\int_{C_R}\prod_{k=1}^n\sqrt{\frac1{1-\frac{z}k}}\,\frac{\sqrt{\pi}\cos\pi z} {\sqrt{z\sin\pi z}}\,dz.$$ So, letting $R\rightarrow\infty$, we arrive at $$\sum_{k=0}^n\sqrt{\binom{n}k} =\frac1{2\pi i}\left[\int_{-\frac12+i\infty}^{-\frac12-i\infty} +\int_{n+\frac12-i\infty}^{n+\frac12+i\infty} \sqrt{g(z)}\,\frac{\pi\cos\pi z}{\sin\pi z}\,dz\right].$$ The integrand is invariant under $z\rightarrow n-z$, hence it suffices to compute (actually estimate) one of the integrals.

This is where I left off. Perhaps someone can complete the argument. [1]: https://en.wikipedia.org/wiki/Residue_theorem

I add the following if it helps with the details with the estimates. $$\sum_{i=0}^n\sqrt{\binom{n}i} =\frac1{\sqrt{\pi}}\int_0^{\infty}\left(n+1-\sum_{k=0}^ne^{-\binom{n}kx^2}\right)\frac{dx}{x^2}.$$

Define the complex-variable function $$g(z)=\frac{\sin\pi z}{\pi z}\prod_{k=1}^n\frac1{1-\frac{z}k}.$$ Note $f(k)=\binom{n}k$ for $k\in\mathbb{Z}^{\geq0}$. Moreover, $\sqrt{g(z)}$ is analytic in the region $-1<\Re(z)<n+1$ since $g(z)$ is zero-free there. By the [Residue Theorem][1], $$\sum_{k=0}^n\sqrt{\binom{n}k} =\frac1{2\pi i}\int_C\sqrt{g(z)}\,\frac{2\pi}{\sin2\pi z}\,dz$$ where $C$ is a closed (positive) contour inside $-1<\Re(z)<n+1$ going around once about each integer $0\leq k\leq n$. Choose $C_R$ as the rectangular along the lines $\Re(z)=-\frac12$, $\Re(z)=n+\frac12$ and $\Im(z)=\pm R$. Thus $$\int_{C_R}\sqrt{g(z)}\,\frac{2\pi}{\sin2\pi z}\,dz =\int_{C_R}\prod_{k=1}^n\sqrt{\frac1{1-\frac{z}k}}\,\frac{\sqrt{\pi}\,dz} {\cos\pi z\,\sqrt{z\sin\pi z}}.$$ So, letting $R\rightarrow\infty$, we arrive at $$\sum_{k=0}^n\sqrt{\binom{n}k} =\frac1{2\pi i}\left[\int_{-\frac12+i\infty}^{-\frac12-i\infty} +\int_{n+\frac12-i\infty}^{n+\frac12+i\infty} \sqrt{g(z)}\,\frac{2\pi}{\sin2\pi z}\,dz\right].$$ The integrand is invariant under $z\rightarrow n-z$, hence it suffices to compute (actually estimate) one of the integrals.

This is where I left off. Perhaps someone can complete the argument. [1]: https://en.wikipedia.org/wiki/Residue_theorem

Not a complete answer yet, but an upper bound and adding some near-evidence to your claim.

By the Cauchy-Schwarz inequality, \begin{align} \left(\sum_{i=0}^n1\cdot \sqrt{\binom{n}i}\right)^2 &\,\,\leq \,\,\left(\sum_{i=0}^n1\right)\cdot\left(\sum_{i=0}^n\binom{n}i\right)=n\cdot 2^n. \end{align} Therefore, $$\sum_{i=0}^n\sqrt{\binom{n}i}\,\,\leq\, \, n^{0.5}\cdot2^{0.5n}.$$

I also add the following if it helps with the details of downsizing $n^{0.5}$ to $n^{0.25}$. $$\sum_{i=0}^n\sqrt{\binom{n}i} =\frac1{\sqrt{\pi}}\int_0^{\infty}\left(n+1-\sum_{k=0}^ne^{-\binom{n}kx^2}\right)\frac{dx}{x^2}.$$

I add the following if it helps with the details with the estimates. $$\sum_{i=0}^n\sqrt{\binom{n}i} =\frac1{\sqrt{\pi}}\int_0^{\infty}\left(n+1-\sum_{k=0}^ne^{-\binom{n}kx^2}\right)\frac{dx}{x^2}.$$