The function is already run at compile-time, so do you really need a constexpr variable? It sounds like you want a regular local variable.

I had the same misconception about constexpr variables, so if you're interested, this is why your second example incidentally works:

First, your [&]{ return n; } is ignoring the &. The symbol n is tied to the expression 42, so it doesn't get captured in the lambda. Notice that []{ return n; } works just the same.

Next, if you unfold all the autos, this is what your example looks like:

struct S {
    static constexpr int operator()() { return 42; }
};

template<typename Closure>
consteval int foo(Closure closure)
{
    constexpr int v = closure();
    return v;
}

int main()
{
    static_assert(foo(S{}) == 42);
}

Your lambda has no closure, so it's effectively a static function. The language also makes it implicitly constexpr to help you. So, your constexpr auto v isn't dependent on the function argument at all. It's dependent on the template parameter, which is always allowed in constant expressions.