From the plot of volatility surface, as maturity goes up, the implied volatility will decrease. Dose it mean that options with the same strike have higher value when maturity is larger. If so why price of large maturity option is often higher, which can contribute to calendar spread?
I'll try to make a guess. Maybe it's due to the properties of the underlying, since we know that GBM will go to almost every point with a sufficient amount of time. The higher the time, the higher the probability of hitting a given price, the higher the value of an option.
An option is hedged by holding some stocks and risk-free bonds.
Option Price = a * stock + (1-a) * risk-free-bonds
where a is the position of the stock needed for hedging in the self-finance portfolio. Therefore, the option price depends on price of the zero coupon bond.
Now, if we have two options, everything the same but time-to-maturity, let's say T1, T2, T2 > T1. The T2 option needs to be hedged with a T2 zero-coupon-bond. However, the T2 bond costs less than the T1 bond because everyone prefers a bond that pays back the money earlier.
Therefore, the price of the option must go up for the T2 option. The relationship can be explained by how Black-Scholes is derived in the first-place (delta continuous hedging).
First of all, IV does not always decrease with maturity. Even with focus on the ATM IV volatility, the term structure can be very different in different circumstances: it can go up, down or be rather bumpy. Nevertheless, in the BS formula the option price is a function of $\sigma(\tau)\sqrt \tau$, so even if $\sigma(\tau)\downarrow$ it may still happen that $\sigma(\tau)\sqrt \tau \uparrow $, otherwise clearly you'll get a negative calendar spread which would be quickly arbitraged by other market participants.