Let $A$ be a finite dimensional algebra. Recall that a module $M$ is called $n$-torsionfree in case $Ext_A^i(D(A),\tau(M))=0$ for all $i=1,...,n$ when $\tau$ denotes the Auslander-Reiten translate. For example for $n=1$ this is equivalent to $M$ being torsionless and for $n=2$ this is equivalent to $M$ being reflexive.
Recall that a module $M$ with minimal injective coresolution $(I_i)$ is said to have dominant dimension at least $n$ in case $I_0,I_1,...,I_{n-1}$ are projective. Let $P$ be an indecomposable projective non-injective module. Let $TF_n$ denote the full subcategory of $n$-torsionfree modules and $Dom_n$ the full subcategory of modules with dominant dimension at least $n$.
Let $0 \rightarrow P \rightarrow X \rightarrow \tau^{-1}(P) \rightarrow 0$ be an almost split sequence and assume that $X$ and $\tau^{-1}(P)$ are $n$-torsionfree for some $n \geq 1$ and each indecomposable projective non-injective module $P$.

> Conjecture: Then the regular module $A$ has dominant dimension at least $n$.

The case $n=2$ follows from a result of Tachikawa and a proof of the conjecture would improve the main result of Tachikawa (see his paper "Reflexive Auslander-Reiten sequences"). The computer gives good evidence that the conjecture is true. 

Here my attempt to prove it by induction for $n \geq 2$ (I look at the case $n=1$ later, but maybe there is a nice proof without induction?):

Assume the conjecture holds for $n-1$. We prove it for $n$ then.
By induction we have that $A$ has dominant dimension at least $n-1$. Showing that $A$ has dominant dimension at least $n$ is equivalent to showing that each indecomposable projective non-injective module $P$ has dominant dimension at least $n$. Now by a result of Auslander and Reiten, $A$ having dominant dimension at least $n-1$ implies that $TF_{l}=Dom_{l}$ for all $l \leq n-1$. Thus $X$ and $N:= \tau^{-1}(P)$ have dominant dimension at least $n-1$. 
In case we can show that $X$ has dominant dimension at least $n$ we are finished since this would imply that $P$ has dominant dimension at least $n$. 
Now $X$ has dominant dimension at least $n$ iff $\Omega^{-(n-1)}(X)$ has domiant dimension at least one, which is equivalent to $\Omega^{-(n-1)}(X)$ being torsionless. That the module $\Omega^{-(n-1)}(X)$ is torsionless is equivalent to $Ext_A^1(D(A),\tau(\Omega^{-(n-1)}(X)))=0$.
The assumption that $X$ is $n$-torsionfree this gives us that $0=Ext_A^n(D(A),\tau(X)) \cong Ext_A^1(D(A) , \Omega^{-(n-1)}(\tau(X))$, which is what we need to show.

Now the result would follow in case we have: 
$Ext_A^1(D(A) , \Omega^{-(n-1)}(\tau(X))=Ext_A^1(D(A),\tau(\Omega^{-(n-1)}(X)))=0$, but I see at the moment no reason why we can "exchange" $\Omega^{n-1}$ with $\tau$ here.

So my question is:

>Do we have: 
$Ext_A^1(D(A) , \Omega^{-(n-1)}(\tau(X))=Ext_A^1(D(A),\tau(\Omega^{-(n-1)}(X)))=0$?