If two people pull on either side of the rope with a force of 65N, why is the tension in the rope 65N?

The rope experiences the same pulling force on both sides, and therefore it is in a state of static equilibrium (i.e. it is at rest consequently has a net force of zero).

A force diagram:

Where #vecF_1# and #vecF_2# are the pulling forces on the rope and #vecT# is the tension force. #R# will represent the rope later.

Here we consider Newton's third law.

Newton's third law states that every force occurs as a member of an action/reaction pair of forces. You may know it as the familiar phrase, "every action has an equal and opposite reaction." Most importantly, two members of an action/reaction pair are equal in magnitude but opposite in direction.

#vecF_(A on B)=-vecF_(B on A)#

In this situation, each person pulls on the rope with #65N# of force, so:

#vecF_(1 on R)=65N# and #vecF_(2 on R)=65N#.

This tells us that the pulling force exerted by each person on the rope is #65N#, and consequently by NIII, the rope exerts an equal but opposite tension force of #65N# on each person.

This is an action reaction pair. Therefore:

#vecF_(R on 2)=vecF_(2 on R)=vecF_(1 on R)=vecF_(R on 1)=65N#

The tension force in the rope, given by #vecF_(R on 1)# and #vecF_(R on 2)#, is therefore #65N#.