Even if $A$ is a finite dimensional algebra over $k$, there may be no bound on the dimension of $\text{End}_A(X)/m$.

Let $Q$ be the Kronecker quiver (i.e., the quiver with two vertices and two arrows from the first vertex to the second), and let $A=kQ$ be its path algebra over $k$.

Suppose $k(\alpha)$ is a simple algebraic field extension of $k$. Let $X$ be the representation of $Q$ over $k$ with $k(\alpha)$ at both vertices, the first arrow acting as the identity, and the second arrow acting as multiplication by $\alpha$. Then $\text{End}_A(X)\cong k(\alpha)$.

So as long as $k$ has simple algebraic extensions of unbounded degree, there is no bound on $\dim_k\left(\text{End}_A(X)/m\right)$ for finite dimensional indecomposable $A$-modules $X$.

Even if $A$ is a finite dimensional $k$-algebra, there may be no bound on the dimension of $\text{End}_A(X)/m$.

Let $Q$ be the Kronecker quiver (i.e., the quiver with two vertices and two arrows from the first vertex to the second), and let $A=kQ$ be its path algebra over $k$.

Suppose $k(\alpha)$ is a simple algebraic field extension of $k$. Let $X$ be the representation of $Q$ over $k$ with $k(\alpha)$ at both vertices, the first arrow acting as the identity, and the second arrow acting as multiplication by $\alpha$. Then $\text{End}_A(X)\cong k(\alpha)$.

So as long as $k$ has simple algebraic extensions of unbounded degree, there is no bound on $\dim_k\left(\text{End}_A(X)/m\right)$ for finite dimensional indecomposable $A$-modules $X$.

Even if $A$ is a finite dimensional algebra over $k$, there may be no bound on the dimensions of $\text{End}_A(X)/m$.

Let $Q$ be the Kronecker quiver (i.e., the quiver with two vertices and two arrows from the first vertex to the second), and let $A=kQ$ be its path algebra over $k$.

Suppose $k(\alpha)$ is a simple algebraic field extension of $k$. Let $X$ be the representation of $Q$ over $k$ with $k(\alpha)$ at both vertices, the first arrow acting as the identity, and the second arrow acting as multiplication by $\alpha$. Then $\text{End}_A(X)\cong k(\alpha)$.

So as long as $k$ has simple algebraic extensions of unbounded degree, there is no bound on $\dim_k\left(\text{End}_A(X)/m\right)$ for finite dimensional indecomposable $A$-modules $X$.

Even if $A$ is a finite dimensional algebra over $k$, there may be no bound on the dimension of $\text{End}_A(X)/m$.

Let $Q$ be the Kronecker quiver (i.e., the quiver with two vertices and two arrows from the first vertex to the second), and let $A=kQ$ be its path algebra over $k$.

Suppose $k(\alpha)$ is a simple algebraic field extension of $k$. Let $X$ be the representation of $Q$ over $k$ with $k(\alpha)$ at both vertices, the first arrow acting as the identity, and the second arrow acting as multiplication by $\alpha$. Then $\text{End}_A(X)\cong k(\alpha)$.

So as long as $k$ has simple algebraic extensions of unbounded degree, there is no bound on $\dim_k\left(\text{End}_A(X)/m\right)$ for finite dimensional indecomposable $A$-modules $X$.