I think this is truly bizarre reasoning, and I'm afraid I have to disagree. The wedge product of 1-forms ##\alpha, \beta## is defined as
$$\alpha \wedge \beta = \alpha \otimes \beta - \beta \otimes \alpha$$
And in general, it's just the antisymmetrized tensor product. So of course the wedge product of the ##n## gradients ##d x^\mu## is a tensor!
I think Carroll has actually managed to confuse himself. I don't have a copy of his book, but I would assume he fixed this issue in his notes during the process of editing them for the book.
If you look at eq. (2.40), Carroll correctly points out that the Levi-Civita symbol (i.e. the object ##\tilde \epsilon_{\mu_1 \ldots \mu_n}## whose entries are always the numbers ##0, \pm 1##) is a tensor density. In other words, ##\tilde \epsilon_{\mu_1 \ldots \mu_n}## is not a covariant object; it's just a bookkeeping device that is useful for writing down certain formulas.
Carroll then considers the object
$$d x^1 \wedge \ldots \wedge dx^n = \frac{1}{n!} \, \tilde \epsilon_{\mu_1 \ldots \mu_n} \, d x^{\mu_1} \wedge \ldots \wedge d x^{\mu_n}$$
and tries to argue what sort of object it is based on how it transforms. In the first line of eq. (2.47), Carroll has it completely right. If we call the components of this object ##A_{\mu_1 \ldots \mu_n}##, then the first line of (2.47) shows
$$A_{\mu_1' \ldots \mu_n'} = A_{\mu_1 \ldots \mu_n} \frac{\partial x^{\mu_1}}{\partial x^{\mu_1'}} \cdots \frac{\partial x^{\mu_n}}{\partial x^{\mu_n'}}$$
which is the transformation law for a tensor! It just so happens that in one coordinate system (the unprimed ##x^\mu##), the components ##A_{\mu_1 \ldots \mu_n}## happen to have values given by
$$A_{\mu_1 \ldots \mu_n} = \tilde \epsilon_{\mu_1 \ldots \mu_n}$$
The second line of eq. (2.47), however, is very misleading, because Carroll has plugged back in ##\tilde \epsilon_{\mu_1' \ldots \mu_n'}##, which we have already agreed is not a covariant object. So the second line of (2.47) is really just reiterating the fact that the Levi-Civita symbol (whose entries are always the numbers ##0, \pm 1##) fails to be a tensor.
But the object ##dx^1 \wedge \ldots \wedge dx^n## is absolutely a tensor. When you change it to another coordinate system ##x^{\mu'}##, it will become
$$dx^1 \wedge \ldots \wedge dx^n = \frac{1}{n!} \, \frac{\partial x^{1}}{\partial x^{\mu_1'}} \cdots \frac{\partial x^{n}}{\partial x^{\mu_n'}} \, dx^{\mu_1'} \wedge \ldots \wedge dx^{\mu_n'}$$
as expected. This notion that it should somehow be a tensor density comes from the false expectation that you should be able to simply replace ##x^\mu \to x^{\mu'}##. Clearly you cannot do that, because the ##x^\mu## are sitting behind partial derivative operators (that is what ##d## is, after all).
If you disagree, let me ask you this: If ##dx^1 \wedge \ldots \wedge dx^n## is a tensor density rather than a proper ##n##-form, then what is
$$dx^1 \wedge \ldots \wedge dx^{n-1} + dx^2 \wedge \ldots \wedge dx^n$$
? (Each term has ##n-1## factors).By the way, the mathematical definition of a tensor is a map from ##T : V^m \times (V^*)^n \to \mathbb{R}## which is function-linear; i.e. for some vectors ##X, Y \in V^m \times (V^*)^n##, we have
$$T(f X + g Y) = f \, T(X) + g \, T(Y)$$
where ##f, g## are functions, not just numbers. You should be able to show that this definition is equivalent to the "transforms as such-and-such" definition. Moreover, it should be obvious that the wedge product of a bunch of one-forms is a tensor.
