Okay, so several people have mentioned that the planet's density will have an effect on its escape velocity. This is true. The total mass of the planet is: M=(4π/3)R³ρ, where ρ is its average density. The escape velocity squared is: v² = 2GM/R = (8πG/3)R²ρ = k²R²ρ, where I've defined k²=8πG/3. So we see that the equation for escape velocity, in terms of its radius and its density, is given by:
v = kR √(ρ)
The acceleration due to gravity is given by g = GM/R²= (4πG/3)Rρ = (k²/2)Rρ
Kip Thorne gives an estimate for Miller's planet's average density: ~10,000 kg/m³, compared with Earth's value of ~5,500 kg/m³. Additionally, we know Miller's planet has 1.2 times the acceleration due to gravity on Earth: g₂=1.2g₁.
R₂ = 1.2R₁ρ₁/ρ₂
This gives us a radius of ~4200 km. The escape velocity can be obtained by plugging this into our first equation, giving us a value of of ~9930 m/s. This is about 90% of Earth's escape velocity, so indeed it is easier to take off from Miller's planet than Earth, however not by much.