After reading the above paper, I guess I have figured out why it is slightly larger than 0.5 without calculating N(d1).
Given that the stock price follows a lognormal distribution, the expected value of the stock price over the current stock price E(St/S0) = exp(µt) where µ is the drift term of geometric Brownian motion. By risk-neural pricing, risk-free rate Rf is used instead of µ, so the E(St/S0) = exp(Rf*t). Before maturity, t is larger than 0, so the expected value is larger than 1. That being said, the expected value of the stock price at maturity with time-to-maturity = t is larger than the current stock price S0. Given that the call option is at the money, the current stock price equals the exercise price. Therefore, the stock price at maturity is expected to be slightly larger than the exercise price, which causes the probability of exercising the option to be slightly larger than 0.5. So the delta of the ATM call option prior to maturity is slightly larger than 0.5.
In simple words, the stock price is expected to grow at a risk-free rate like how forwards are priced but using continuously compounded interest rate instead. So the delta is > 0.5.
Correct me if I am wrong..