Let me write $c=1/2+\delta$. Since $$P_n(\delta)\equiv \mathbb{P}\left[\sum_{i=1}^n X_i \leq n \cdot c\right]$$ satisfies $P_n(0)=1/2$ and $P_n(1/2)=1$, irrespective of $n$, in order to demonstrate that $P_n(\delta)$ increases with $n$ for $0<\delta<1/2$ we need to prove
(I) that the slope of $P_n(\delta)$ at $\delta=0$ increases with $n$ and
(II) that $P_n(\delta)$ is a concave function of $\delta$.
It is convenient to work with the characteristic function of the Irwin-Hall distribution. I find the principal value integral $$\mathbb{P}\left[\sum_{i=1}^n X_i \leq n \cdot c\right]=\frac{1}{2}-\frac{1}{2\pi}\int_{-\infty}^\infty dt\,\frac{1}{(it)^{n+1}}\left(e^{it(1-c)}-e^{-ict}\right)^n,$$ which can be rewritten as $$P_n(\delta)=\frac{1}{2}+\frac{1}{\pi}\int_{0}^\infty dt\,\sin(2nt\delta)\,\frac{\sin^n t}{t^{n+1}}$$
(I) Proof that the slope at $\delta=0$ increases with $n$.
For small $\delta$ the integral evaluates to
$$P_n(\delta)=\tfrac{1}{2}+2nC_n\delta+{\cal O}(\delta^3)$$
with the coefficient
$$C_n=\frac{1}{\pi}\int_0^\infty dt\,\frac{\sin^n t}{t^n}$$
This integral over the $n$-th power of the sinc function is well-studied, we need to show that it decreases more slowly than $1/n$. For small $n$ this can be checked by explicit calculation, for large $n$ it follows from the asymptotic decay $C_n\propto 1/\sqrt n$. So the slope $2nC_n$ is indeed an increasing function of $n$.
(I) Proof of concavity --- in progress.
We need to show that the second derivative $$ P''_n(\delta)=-\frac{4n^2}{\pi}\int_{0}^\infty dt\,\sin(2nt\delta)\,\frac{\sin^n t}{t^{n-1}}$$ is negative for $0<\delta<1/2$.