The expected number of values hit is asymptotic to $\sqrt{n\log n}$.

Start with Stirling's formula:
$$ P(n,k) := 2^{-n}\binom{n}{n/2+k} = \sqrt{\frac{2}{\pi n}} e^{-2k^2/n} (1 + O(1/n)), $$
provided $k$ is not too large (smaller than $n^{1/3}$ is more than enough).

The probability that value $n/2+k$ is hit is
$$P(n,k)=1 - \biggl(1 - 2^{-n}\binom{n}{n/2+k}\biggr)^n.$$
Now scale by $k=x\sqrt{n\log n}$.  Dropping the error terms for simplicity, and using $(1-z)^n=e^{-zn+O(z^2n)}$ we have
$$P(n,k) \approx P_0(n,x) := 1 - \exp\biggl( -\frac 2\pi n^{1/2-2x^2} \biggr).$$

The function $P_0(n,x)$ has a plateau shape with tiny tails.  As $n\to\infty$, we have
$$ P_0(n,x) ~~\begin{cases} \to 1, & |x|<1/2\\
                          = 1 - e^{-\sqrt{2/\pi}}, &|x|=1/2\\
                          \to 0, & |x|>1/2,
\end{cases} $$
where in the first and third cases convergence is very rapid.
Thus we have
$$\int_{-\infty}^\infty P_0(n,x)\,dx \to 1 ~\text{ as $n\to\infty$}.$$

The sum and the integral are asymptotically equal (which in this case is easy to see since the function rapidly converges to a step function), so we conclude that
$$\sum_{k=-n/2}^{n/2} P(n,k) \sim \sqrt{n\log n}.$$
In detail, for any $\epsilon>0$, values in $[n/2-(1/2-\epsilon)\sqrt{n\log n},n/2+(1/2-\epsilon)\sqrt{n\log n}]$ almost certainly appear and values outside $[n/2-(1/2+\epsilon)\sqrt{n\log n},n/2+(1/2+\epsilon)\sqrt{n\log n}]$ almost certainly don't appear.

All this could be done with more precision.

ADDED: Working directly, if $|k|\le \frac12\sqrt{n\log n}-\sqrt{n}$ then $n/2+k$ almost certainly occurs, while if $|k|\ge \frac12\sqrt{n\log n}+\sqrt{n}$ then $n/2+k$ almost certainly does not occur. Doing this with error estimates shows that the expected number of values hit is $\sqrt{n\log n}+O(\sqrt n)$.