Although we can define the momentum as a self-adjoint operator in $L^2[0,1]$ as you proposed, I think it's rather artificial to think about it as having relation to momentum in the case of $L^2(\Bbb{R})$. Realize that the operator $p_1$ with domain
$D(p_1)=\{\psi\in\mathcal{H}^1[0,1]\,|\,\psi(1)=\psi(0)\}$, is related to spatial translations via the unitary group $U(t)=\exp(-itp_1)$, whose action is
$$(U(t)\psi)(x)=\psi[x-t\pmod{1}],$$
so it's about a particle in a torus, not in an infinite square-well. Different values of $\alpha$, just give different phases to the wavefunction when it reaches the border and goes to the other side.
So, in my opinion, these operators are not actually related to the momentum as usually conceived. The idea of an infinite square-well does not allow spatial translations, so there's no self adjoint operator associated to a unitary translation group in this case. This happens for example in the case of a particle in the postive real line $\Bbb{R}_+$. In this case, the space $L^2[0,\infty)$ allows only translations to the right, not to the left, so you can not have a self-adjoint operator associated to a unitary group of translations. In this case, the operator $p=-i\dfrac{d}{dx}$ has no self-adjoint extensions, for any initial domain, although it is symmetric. For a particle in a box, we can think in the same way. There's no operator associated to spatial translations, because there is no spatial translations allowed.
It's also important to note that the hamiltonian $H$ in this case is given by the Friedrich extension of $$p_0^2=-\frac{d^2}{dx^2}\\
\mathcal{D}(p_0^2)=\{\psi\in\mathcal{H}^2[0,1]\,|\,\psi(0)=\psi'(0)=0=\psi'(1)=\psi(1)\}$$
$H$ cannot be the square of any $p_\alpha$, since the domains do not match.
Edit: As pointed out by @jjcale, one way to take the momentum in this case should be $p=\sqrt{H}$, but clearly, the action of $p$ can't be a derivative, because it has the same eigenfunctions of $H$, which are of the form $\psi_k(x)=\sin \pi kx$. This ilustrates the fact that it's not related to spatial translations as stated above.
Edit 2: There's is a proof that the Friedrich extension is the one with Dirichilet boundary conditions in Simon's Vol. II, section X.3.
The domains defined by the spectral theorem are indeed $\{\psi: p_\alpha\psi\in\mathcal{D}(p_\alpha)\}$. To see this, realize that in this case, since the spectrum is purely point, by the spectral theorem, we have
$$p_\alpha=\sum_{n\in \Bbb{Z}}\lambda_{\alpha,n}P_n,$$
where $\lambda_{\alpha,n}$ are the eigenvalues associated to the normalized eigenvectors $\psi_{\alpha,n}$, and $P_n=\psi_n\langle\psi_n,\cdot\rangle$ are the projections in each eigenspace. The domain $\mathcal{D}(p_\alpha)$ is then given by the vectors $\xi$, such that
$$\sum_{n\in \Bbb{Z}}|\lambda_{\alpha,n}|^2\|P_n\xi\|^2=\sum_{n\in \Bbb{Z}}|\lambda_{\alpha,n}|^2|\langle\psi_n,\xi\rangle|^2<+\infty$$
Also, $\xi\in\mathcal{D}(p_\alpha^2)$ iff
$$\sum_{n\in \Bbb{Z}}|\lambda_{\alpha,n}|^4\|P_n\xi\|^2=\sum_{n\in \Bbb{Z}}|\lambda_{\alpha,n}|^4|\langle\psi_n,\xi\rangle|^2<+\infty$$
But then, $p_\alpha\xi$ is such
$$\sum_{n\in \Bbb{Z}}|\lambda_{\alpha,n}|^2\|P_np_\alpha\xi\|^2=\sum_{n\in \Bbb{Z}}|\lambda_{\alpha,n}|^2|\langle\psi_n,p_\alpha\xi\rangle|^2=
\sum_{n\in \Bbb{Z}}|\lambda_{\alpha,n}|^2|\langle p_\alpha\psi_n,\xi\rangle|^2=
\sum_{n\in \Bbb{Z}}|\lambda_{\alpha,n}|^4|\langle\psi_n,\xi\rangle|^2<+\infty$$
So, $\mathcal{D}(p_\alpha^2)= \{\psi: p_\alpha\psi\in\mathcal{D}(p_\alpha)\}$.
In QM the position operator and the Hilbert space of a particle are defined contextually: The Hilbert space is $L^2(\mathbb R^3, d^3x)$ and the operator position along $x_k$ is defined, in that space, as $(X_k\psi)(x):= x_k \psi(x)$ with the obvious domain.
You can adopt a more abstract viewpoint if you simultaneously define the momentum and the position operator satisfying CCR, exploiting the so-called Stone-von Neumann theorem. If the representation of the CCR is irreducible (and some technical requirements hold) then the Hilbert space is unitarily isomorphic to $L^2(\mathbb R^3, d^3x)$ and the position and momentum operators become the standard ones under the mentioned isomorphism.
Another more sophisticated approach relies upon the notion of (Mackey's) system of imprimitivity, where the momentum operator is the generator of spatial translations of the three position momenta.
Best Answer
In a complex Hilbert space ${\cal H}$, an operator $A: D(A) \to {\cal H}$, with $D(A)\subset {\cal H}$ a (not necessarily dense) subspace, is said to be Hermitian if $$\langle A \psi | \phi \rangle = \langle \psi| A \phi\rangle \quad \forall \psi, \phi \in D(A)\:. \quad (1)$$ It seems to be worth stressing that, to check (1), it is not necessary to exploit the definition of adjoint operator, $A^\dagger$ that, generally, does not exist when $D(A)$ is not dense.
If $D(A)$ is dense, the Hermitian operator $A$ is said to be symmetric.
In your case(s) $A:=T_n$ and $D(T_n)= S({\mathbb R})$, the Hilbert space ${\cal H}$ being $L^2(\mathbb R)$.
Just using (a) integration by parts, (b) the definition of scalar product in $L^2(\mathbb R)$ and (c) the fact that the elements of $S(\mathbb R)$ rapidly vanishes for $|x|\to \infty$ with all derivatives, you immediately establish that (1) is valid for $T_n$.
Actually these operators are symmetric because $S(\mathbb R)$ is dense in $L^2(\mathbb R)$.
Existence of self-adjoint extensions can be studied examining deficiency indices of symmetric operators. It is not necessary to prove that $T_n$ is closed, even if this condition is assumed in several textbooks as Reed and Simon, as stated in Dunford Schwartz books (vol II Corollary 13 ch. XII.4.13). However, if $n$ is even, self-adjoint extensions do exist in view of the fact that $T_n$ commute with the anti unitary involutive operator given by the complex conjugation of Fourier transforms (exploiting Theorem 18 ch. XII.4.13) which establishes that the deficiency indices coincide.