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 <A HREF="https://en.wikipedia.org/wiki/Irwin–Hall_distribution">Irwin-Hall distribution</A>. 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
$$P_n(\delta)=\frac{1}{2}+\frac{1}{\pi}\int_{0}^\infty dt\,\sin(2nt\delta)\,\frac{\sin^n t}{t^{n+1}}$$

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**(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 <A HREF="http://mathworld.wolfram.com/SincFunction.html">well-studied</A>, 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 $n$.

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**(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 $0<\delta<1/2$.