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 $n^{0.25}$. $$\sum_{i=0}^n\sqrt{\binom{n}i} =\frac1{\sqrt{\pi}}\int_0^{\infty}\left(n-\sum_{k=0}^ne^{-\binom{n}kx^2}\right)\frac{dx}{x^2}.$$