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Let me write $c=1/2+\delta$ and define $$P_n(\delta)\equiv \mathbb{P}\left[\sum_{i=1}^n X_i \leq n \cdot c\right].$$ Note that $P_n(0)=1/2$ and $P_n(1/2)=1$, irrespective of $n$.

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 $$P_n(\delta)=\frac{1}{2}+\frac{1}{\pi}\int_{0}^\infty dt\,\sin(2nt\delta)\,\frac{\sin^n t}{t^{n+1}}.$$

The plot shows $P_n(\delta)$ for $n=1,2,3,4$, from bottom curve to top curve. We need to demonstrate that this ordering of $P_n(\delta)$ with increasing $n$ holds for all $n$, so $P_n(\delta)$ increases with $n$ for all $\delta\in(0,1/2)$. I will attempt to prove this in several steps.

(I) $P_n(\delta)$ increases with $n$ near $\delta=0$.

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 $P'_n(0)=2nC_n\propto \sqrt n$ is indeed an increasing function of $n$.

(II) $P_n(\delta)$ increases with $n$ near $\delta=1/2$.

This follows from a similar expansion around $\delta=1/2$, which shows that the first nonzero $p$-th order derivative $P_n^{(p)}(1/2)$ occurs for $p=n$. So near $\delta=1/2$ the function expands as

$$P_n(\delta)=1-(-1)^n A_n(\delta-1/2)^n+{\cal O}(\delta-1/2)^{n+1},\;\;A_n>0,$$

hence $P_n(\delta)$ increases with $n$ for $\delta$ just below $1/2$.

(III) Large-$n$ asymptotics

For $n\gg 1$ the sinc integral can be evaluated in closed form by means of the limit $$\lim_{n\rightarrow\infty}\left(\frac{\sin(t/\sqrt n)}{t/\sqrt n}\right)^n=\exp(-t^2/6)$$ so that $P_n(\delta)$ becomes asymptotically $$P_n(\delta)\rightarrow\frac{1}{2}+\frac{1}{\pi}\int_0^\infty dt\,\sin(2t\delta\sqrt n)\frac{1}{t}\exp(-t^2/6)=\frac{1}{2}+\frac{1}{2}\,{\rm Erf}\,(\delta\sqrt{6n}),$$ which has indeed a monotonically increasing $n$-dependence for each $\delta\in(0,1/2)$.

Note that we recover the $\sqrt n$ slope at $\delta=0$, $$P'_n(0)\rightarrow \sqrt{6n/\pi}\;\;\text{for}\;\;n\rightarrow\infty.$$ At $\delta=1$ the deviation from the exact limit $P_n(1/2)=1$ is exponentially small, $\propto\exp(-3n/2)/\sqrt n$.

The error function asymptotics is remarkably accurate already for small $n$, see the plot below for $n=3$ and $n=4$. For larger $n$ the two curves are almost indistinguishable.

Bottom line.

I would think that the finite-$n$ analytics near $\delta=0$ and $\delta=1/2$, together with the large-$n$ asymptotics for the whole interval $\delta\in(0,1/2)$, goes a long way towards a proof of the monotonic increase with $n$ of $P_n(\delta)$, although some further estimation of the error in the asymptotics is needed to complete the proof. I am surprised by the fact that the large-$n$ asymptotics is so accurate already for $n=4$.