Dot product (of coordinate vectors) is indeed dependent on coordinates. For instance, the vector
$$
e_1
$$
has coordinate vector $v = (1,0,0)$
in the basis $e_1, e_2, e_3$ which has $$v \cdot v = 1,$$
but the same vector, expressed in the basis
$$\frac{1}{2} e_1, e_2, e_3$$
has coordinate vector $$
w =(2, 0, 0),
$$
and $w\cdot w = 4$.
post-comment additions:
To explain the other problem you're having:
When you write "Let
$$ \textbf{v}=R\textbf{e}_r+\Theta\textbf{e}_{\theta}+Z\textbf{e}_z\in \mathbb{R}^3 \ldots,$$
you've already gone off the rails. Because the basis you've written at the top of your post is the basis for the space of "vectors based at the point whose coordinates, in the polar system, are $(r, \theta, z)$." It's not a basis for $\mathbb R^3$, because in the context of $\mathbb R^3$, the values $r$ and $\theta$ aren't even defined!
This'll become obvious once I make things concrete: Let's the the vector $v$ where $R = 2, \Theta = \frac{\pi}{2}, Z = 1$. In the expression for $e_r$ that you plugged in, what's the value of $\theta$? You seem to have plugged in $\Theta$, but why? For the basis to make sense, you need the $(r, \theta, z)$ coordinates of the point at which you're using it as a basis, but you don't have those...so you've used the nearest thing, typographically, as a substitute. There's really no justification for that.
Some Gratuitous Advice
A question for you: When you replaced the lower-case $\theta$ and $r$ with their upper-case versions, did something in the back of your mind say, "Hey, wait a minute...these are actually different!"? And did you then perhaps say "Yeah, but they're the only "r" and "theta" I can see in the formulas I've got, so I guess I have to use them!"? Because that voice in your head was the warning that you were doing something wrong, and needed a deeper understanding before proceeding.
I spend a good deal of time programming, and I find debugging about 10 times as hard as programming. That's pretty much true for math as well, and listening to that little voice is part of the way to avoid debugging (in both contexts).
Any orthonormal set of vectors ${u_1, u_2, u_3}$ will satisfy
$u_1\times u_2 = u_3$ or $-u_3$, and
$u_2\times u_3 = u_1$ or $-u_1$ and
$u_1\times u_3 = u_3$ or $-u_3$.
Your coordinates in this case form an orthonormal basis, and they are right handed, so they will obey the rules above with all positive signs. Alternatively, you could just right out each of the vectors in cyclindrical coordinates and use the cross products of the standard basis to arrive at an answer.
Best Answer
In cylindrical coordinates, you have the unit vectors
$$\begin{cases}\hat r = \frac{x\,\hat\imath+y\,\hat \jmath}{r} = \frac{r\cos(\theta)\,\hat \imath + r\sin(\theta)\,\hat \jmath}{r} = \cos(\theta)\,\hat \imath + \sin(\theta)\,\hat \jmath \\ \hat z = \hat k \\ \hat\theta = \hat z \times \hat r = -\sin(\theta)\,\hat\imath + \cos(\theta)\,\hat \jmath\end{cases}$$
Solving for the Cartesian unit vectors gives
$$\begin{cases}\hat\imath = \cos(\theta)\,\hat r - \sin(\theta)\,\hat\theta \\ \hat\jmath = \sin(\theta)\,\hat r + \cos(\theta)\,\hat\theta \\ \hat k = \hat z\end{cases}$$
So if $\hat x = x_1\,\hat\imath + x_2\,\hat\jmath + x_3\,\hat k$, where $x_1=r\cos(\theta)$ and $x_2=r\sin(\theta)$, its cylindrical coordinate representation would be
$$\hat x = (x_1\cos(\theta)+x_2\sin(\theta))\,\hat r + (x_2\cos(\theta)-x_1\sin(\theta))\,\hat\theta + x_3\,\hat z$$
$-\hat x$ is the same vector, but $\theta$ is replaced with $\theta\pm\pi$, which gives $\cos(\theta\pm\pi)=-\cos(\theta)$ and $\sin(\theta\pm\pi)=-\sin(\theta)$; ultimately the same effect as multiplying $\hat x$ by $-1$.
Then the cross product of $\hat x$ and $-\hat x$ is
$$\hat x\times(-\hat x) = \begin{vmatrix} \hat r & \hat \theta & \hat z \\ x_1\cos(\theta)+x_2\sin(\theta) & x_2\cos(\theta)-x_1\sin(\theta) & x_3 \\ -x_1\cos(\theta)-x_2\sin(\theta) & -x_2\cos(\theta)+x_1\sin(\theta) & -x_3 \end{vmatrix} = \hat 0$$
since the second and third rows are linearly dependent.