(Too long for a comment).

Here is a way to get $\ge c \sqrt{n}$ for some constant $c$: First pick $x$ uniformly at random from the sphere and consider $\mathbf{E}|\langle x,u_1 \rangle|$. We can assume the first vector of the basis is $u_1$ and form the rest of the orthonormal basis. Then the expected value is just the absolute value of the first coordinate $|x_1|$.

To calculate this, we note that we can generate a random vector by taking a random gaussian and normalizing it. This means that

$$\mathbf{E}|\langle x,u_1 \rangle| = \int_0^{\infty} \mathbf{P}(|x_1| \ge t) \ dt \approx \int_0^{\infty} \mathbf{P}(g \ge t \sqrt{n})\ dt $$
where $g$ is a standard normal random variable. In the approximation step, we use strong concentration of chi-squared random variables to say the norm of a random gaussian vector concentrates around $\sqrt{n}$ (the details need to be spelled out but they should be straightforward). Finally, the tail of the gaussian tells us that $\mathbf{P}(g \ge t \sqrt{n}) \le \exp(-t^2n)$ so the integral evaluates to $c/\sqrt{n}$ for some fixed constant $c$.

Since the expected value is at least $c \sqrt{n}$, this tells us that there exists a $x$ for which the bound holds.