7. Given the inequality about $x$
$$
\frac{\mathrm{e}^{x}}{x^{3}}-x-a \ln x \geqslant 1
$$

holds for any $x \in(1,+\infty)$. Then the range of the real number $a$ is
7. $a \leqslant-3$.
Notice that, $\frac{\mathrm{e}^{x}}{x^{3}}=\frac{\mathrm{e}^{x}}{\mathrm{e}^{\ln x^{3}}}=\mathrm{e}^{x-3 \ln x}$.
By the well-known conclusion $\mathrm{e}^{x} \geqslant x+1$, we know
$$
\mathrm{e}^{x-3 \ln x} \geqslant 1+x-3 \ln x.
$$

When $a \leqslant-3$,
$$
\begin{array}{l}
\frac{\mathrm{e}^{x}}{x^{3}}=\mathrm{e}^{x-3 \ln x} \geqslant 1+x-3 \ln x \\
\geqslant 1+x+a \ln x(x>1),
\end{array}
$$

the original inequality always holds.
When $a>-3$, let
$$
\begin{array}{l}
g(x)=x-3 \ln x(x>1) \\
\Rightarrow g^{\prime}(x)=\frac{x-3}{x}.
\end{array}
$$

Thus, $g(x)_{\text {min }}=g(3)=3-3 \ln 30$ makes $g\left(x_{0}\right)=0$, i.e.,
$$
\mathrm{e}^{x_{0}-3 \ln x_{0}}=1,
$$

and at this point,
$$
\begin{array}{l}
x_{0}+a \ln x_{0}+1 \\
=\left(x_{0}-3 \ln x_{0}+1\right)+(a+3) \ln x_{0}>1,
\end{array}
$$

the original inequality does not hold.
In summary, $a \leqslant-3$.