Is nameof() evaluated at compile-time?

Yes. nameof() is evaluated at compile-time. Looking at the latest version of the specs:

The nameof expression is a constant. In all cases, nameof(…) is evaluated at compile-time to produce a string. Its argument is not evaluated at runtime, and is considered unreachable code (however it does not emit an “unreachable code” warning).

From nameof operator – v5

You can see that with this TryRoslyn example where this:

public class Foo
{
    public void Bar()
    {
        Console.WriteLine(nameof(Foo));
    }
}

Is compiled and decompiled into this:

public class Foo
{
    public void Bar()
    {
        Console.WriteLine("Foo");
    }
}

Its run-time equivalent is:

public class Foo
{
    public void Bar()
    {
        Console.WriteLine(typeof(Foo).Name);
    }
}

---

As was mentioned in the comments, that means that when you use nameof on type parameters in a generic type, don’t expect to get the name of the actual dynamic type used as a type parameter instead of just the type parameter’s name. So this:

public class Foo
{
    public void Bar<T>()
    {
        Console.WriteLine(nameof(T));
    }
}

Will become this:

public class Foo
{
    public void Bar<T>()
    {
        Console.WriteLine("T");
    }
}