Smoothness of Hecke algebras

First I will introduce some notation and definitions.

Fix a level $N$ (take $N=1$ if it makes things easier) and a prime $p$. Let $k$ be a finite field of characteristic $p$ and let $\mathcal{C}$ be the category of complete noetherian local rings $(R, \mathfrak{m}_R)$ together with an isomorphism $R/\mathfrak{m}_R \cong k$. Let $\Lambda$ be an artinian object of $\mathcal{C}$ (I'm interested in cases where $\Lambda = \mathcal{O}_K / \pi^n$ where $K$ is a finite extension of $\mathbb{Q}_p$ with uniformiser $\pi$). Let $S_k(\mathbb{Z}_p)$ denote the set of $q$-expansions of cuspforms of level $N$ and coefficients in $\mathbb{Z}_p$, and for a $\mathbb{Z}_p$-algebra $R$ let $S_k(R) := S_k(\mathbb{Z}_p)\otimes_{\mathbb{Z}_p}R$. Let $S(R) = \sum_k S_k(R)$, and let $\mathbb{T}(R)$ denote the $R$-subalgebra of $\mbox{End}(S(R))$ generated by the Hecke operators $T_n$ for all $n$ such that $(n,p)=1$.

Now if I understand things correctly, it follows from deformation theory of Galois (pseudo-)representations that $\mathbb{T}(\Lambda)$ is a complete noetherian semilocal ring. The maximal ideals correspond to mod $p$ modular representations. Let $\mathfrak{m}$ be a maximal ideal.

My question is the following: can we under some conditions assert that $\mathbb{T}(\Lambda)_\mathfrak{m}$ is smooth or $\mathfrak{m}$-smooth as a $\Lambda$-algebra for all $\Lambda$, or at least for all the rings $\Lambda$ I'm interested in (described above) ?

What I know:
Please feel free to correct me if I'm mixing things up in the following.

Let $\mathbb{T} = \mathbb{T}(W(k))$. Let $\mathcal{R}$ be the universal deformation ring attached to the residual representation corresponding to $\mathfrak{m}$. If the deformation problem is unobstructed, then I know that $\mathcal{R}$ will be a power series ring and therefore smooth. Moreover the restriction of the deformation functor to the category of $\Lambda$-algebras is represented by $R_\Lambda = \mathcal{R}\mathbin{\hat\otimes} \Lambda$ and is therefore smooth again. Generally we have a surjective homomorphism $\mathcal{R}\twoheadrightarrow\mathbb{T}$ but under some conditions we have $\mathcal{R} \cong\mathbb{T}$.

However it seems to me that even if we have $\mathcal{R} \cong\mathbb{T}$, we might not have that $\mathcal{R}_\Lambda \cong \mathbb{T}(\Lambda)$, where $\mathbb{T}_\Lambda$ is as defined above. For an example there is the following paper by Bellaiche and Khare http://people.brandeis.edu/~jbellaic/preprint/Heckealgebra4.pdf

In the cases they consider, $\mathbb{T}(k)_\mathfrak{m}$ (which they call $A_{\overline{\rho}}$) is not the reduction of $\mathbb{T}$ modulo $p$, but a quotient of that modulo some element in the maximal ideal. Actually, under some conditions (e.g. the deformation problem is unobstructed) they identity $\mathbb{T}(k)_\mathfrak{m}$ with the universal ring of pseudo-deformations to $k$-algebras with constant determinant. In these cases, we still can deduced that $\mathbb{T}(k)_\mathfrak{m}$ is smooth.

I would appreciate any hints or references on this question. Thanks.

It follows from deformation theory of Galois (pseudo-)representations that $\mathbb T(\Lambda)$ is a complete noetherian semilocal ring. The maximal ideals correspond to mod $p$ modular representations.
The fact that $\mathbb T(\Lambda)$ is a semi-local ring follows from the existence of the natural surjective map $$\mathbb T(\mathcal O_K)/\pi^{n}\longrightarrow \mathbb T(\Lambda)$$ and the fact that system of Hecke eigenvalues (otherwise known as eigenforms) are in finite numbers (not a priori from arguments with pseudo-representations).
Moreover, for stupid reasons of compatibility of coefficients, the maximal ideals of $\mathbb T(\Lambda)$ cannot correspond to mod $p$ modular representations. Now maybe you meant to write "correspond to modular representations with coefficients in $\Lambda$" however, as in your setting the coefficient ring is artinian but not a field, it is not obviously true that such a statement hold: maybe the pseudo-representation $t:G_{\mathbb Q}\longrightarrow \mathbb T(\Lambda)_{\mathfrak m}$ sending $\operatorname{Fr}(\ell)$ to $T(\ell)$ does not lift to an actual representation $\rho_{\mathfrak m}$ with coefficients in $\Lambda_{\mathfrak m}$. Conversely, a single residual representation $\bar{\rho}$ can correspond to several congruent pseudo-representations with coefficients in $\mathbb T(\Lambda)$.
Regarding your actual question, I think the only known line of attack of such problems at present is the one you describe, that is to say the one of Nicolas-Serre-Bellaïche-Khare: either find an explicit description (totally hopeless in your case) or assume that $R(\bar{\rho})=\mathbb T_{\bar{\rho}}$ is smooth and try to relate $\mathbb T_{\bar{\rho}}\otimes_{\mathcal O}\mathcal O/\pi^n$ to $\mathbb T(\Lambda)_{\mathfrak m}$. As I indicated above, this seems to require much clarification as $\mathfrak m$ and $\bar{\rho}$ are not in obvious correspondence anymore (and in particular, I would certainly not bet that unobstructed implies smooth in your setting; I don't see how to rule out the possibility of many congruences between pseudo-representations destroying the smoothness when one passe from $\mathbb T(\mathcal O/\pi)$ to $\mathbb T(\mathcal O/\pi^n)$).