Setting function's prototype to a private variable

This does not work because you are overriding the prototype. When you call new Outer, then this will inherit from the current Outer.prototype. You are then overriding Object.prototype inside the function, but it does not affect the currently created instance, only future ones.
You'd have to extend the existing prototype.

But if you want hide certain functions, use a self-invoking function:

var Outer = (function() {

    function Extensions() {
    }
    Extensions.protoype.Greet = function () {
        alert(this.Greetings);
    }

    function Inner() {
        Extensions.call(this);
        this.Greetings = "Take me to your leader.";
    }
    Inner.prototype = Object.create(Extensions.prototype);
    Inner.prototype.constructor = Inner;

    function Outer() {
        Extensions.call(this);
        this.Greetings = "Greetings!";
    }
    Outer.prototype = Object.create(Extensions.prototype);
    Outer.prototype.constructor = Outer;

    Outer.prototype.getInner = function() {
        return new Inner();
    };

    return Outer;

}());

Here, Extensions is defined inside the immediately executed function. Therefore it is not accessible from the outside. Only Outer can be accessed since you return it from that function.

See this answer for a very good explanation of inheritance done right in JavaScript.