We have \begin{equation*} T(4,1,x)\sim t(x):=e^{-x}/x^3 \tag{1}\label{1} \end{equation*} as $x\to\infty$.
Indeed, from formulas (29) and (30) of the linked paper we get $$T(4,1,x)=\frac{g_2(x)-2 g_1(x) \ln x+e^{-x} \ln^2 x}{2 x},$$ where $$g_k(x):=\int_x^\infty\ln^k u\,e^{-u}\,du.$$ Using now repeatedly integration by parts to obtain enough terms of the asymptotics of $g_1(x)$ and $g_2(x)$, we get \eqref{1} -- see details on this at the end of this answer.
Here is the graph $\{(x,\frac{T(4,1,x)}{t(x)})\colon20\le x\le200\}$:
Details: Let $l$ denote a function of the form $\ln^p$ for some real $p$ or any of its derivatives, of any order. Integrating by parts, for all real $x>1$ we have \begin{equation*} g(x):=\int_x^\infty l(u) e^{-u}\,du=l(x)e^{-x}+\int_x^\infty l'(u) e^{-u}\,du \end{equation*} and hence, by induction, for any natural $k$, \begin{equation*} \begin{aligned} g(x)&=(l(x)+l'(x)+\cdots+l^{(k-1)}(x))e^{-x}+\int_x^\infty l^{(k)}(u) e^{-u}\,du \\ &=(l(x)+l'(x)+\cdots+l^{(k-1)}(x)+l^{(k)}(x)(1+o(1)))e^{-x} \end{aligned} \tag{2}\label{2} \end{equation*} (as $x\to\infty$); the latter equality in \eqref{2} follows by the l'Hospital rule, since $(\ln^p)^{(k)}(x)\asymp x^{-k}\ln^{p-1}x$.
So, by \eqref{2} with $k=3$ and $g\in\{g_1,g_2\}$, \begin{equation*} e^x g_1(x)=\ln x+\frac{1}{x}-\frac{1}{x^2}+\frac{2+o(1)}{x^3}, \end{equation*} \begin{equation*} e^x g_2(x)=\ln^2 x+\frac{2 \ln x}{x}-\frac{2 \ln x}{x^2}+\frac{2}{x^2} +\frac{(4+o(1)) \ln x}{x^3}, \end{equation*} \begin{equation*} e^x T(4,1,x)=\frac{x+o(\ln x)}{x^4}\sim 1/x^3. \quad\Box \end{equation*}