It's always non-trivial to give a definite negative answer to this kind of question. However, the following methods, which tend to be used in cases when closed form answers are obtained, don't seem to lead to anything useful.

• Explicit summation over $n$ with a finite upper limit. Knowing an explicit formula for the finite sum and taking the upper limit to infinity would give the desired series sum. There are methods that can check whether there exists an anti-difference (like an anti-derivative, but for sums instead of integrals) in a large class of functions, like those of hypergeometric type, e.g., Zeilberger's algorithm. Such algorithms are actually already implemented in Mathematica and Maple. So, since asking one of these computer algebra systems doesn't yield an answer, this method probably doesn't work.

• Representation as a sum of residues. Is there a function $g(z)$ and a contour in the complex $z$-plane such that the series equals the sum of residues of $g(z)$ inside this contour? Here are a couple of candidates, $g(z) = -\frac{\pi}{\sin(\pi z)} \frac{\psi(z)}{z} \log z$ or $g(z) = -\frac{1}{2}(\psi(\frac{-z+1}{2})-\psi(\frac{-z}{2})) \frac{\psi(z)}{z} \log z$, with the contour in both cases hugging the positive real axis. The simplification in this representation could come from the ability to deform the contour to encircle an alternative set of poles, with simpler structure, without changing the value of the integral. In this case, the complicated structure of the poles that appear on the negative real axis and the necessary logarithmic branch cut starting at $z=0$, does not seem to lead to any simplification.

• Fourier series and Parseval's identity. This is the method that is often used to compute the sum $\sum_{n=1}^\infty \frac{1}{n^2}$. The summand is split into a product of Fourier coefficients of two known functions and the sum is transformed into an integral of the product of these functions. (One can also use a slight variation of the method with power series instead of Fourier series, as shown here.) While the integral in question can be factored in multiple ways that give Fourier series of known functions (e.g., with Mathematica's help). Their product doesn't seem to be of the type that would have a closed form itself.

Since these methods don't seem to be succeeding, it's likely that there is no answer in commonly used closed forms.

It's always non-trivial to give a definite negative answer to this kind of question. However, the following methods, which tend to be used in cases when closed form answers are obtained, don't seem to lead to anything useful.

• Explicit summation over $n$ with a finite upper limit. Knowing an explicit formula for the finite sum and taking the upper limit to infinity would give the desired series sum. There are methods that can check whether there exists an anti-difference (like an anti-derivative, but for sums instead of integrals) in a large class of functions, like those of hypergeometric type, e.g., Zeilberger's algorithm. Such algorithms are actually already implemented in Mathematica and Maple. So, since asking one of these computer algebra systems doesn't yield an answer, this method probably doesn't work.

• Representation as a sum of residues. Is there a function $g(z)$ and a contour in the complex $z$-plane such that the series equals the sum of residues of $g(z)$ inside this contour? Here are a couple of candidates, $g(z) = -\frac{\pi}{\sin(\pi z)} \frac{\psi(z)}{z} \log z$ or $g(z) = -\frac{1}{2}(\psi(\frac{-z+1}{2})-\psi(\frac{-z}{2})) \frac{\psi(z)}{z} \log z$, with the contour in both cases hugging the positive real axis. The simplification in this representation could come from the ability to deform the contour to encircle an alternative set of poles, with simpler structure, without changing the value of the integral. In this case, the complicated structure of the poles that appear on the negative real axis and the necessary logarithmic branch cut starting at $z=0$, does not seem to lead to any simplification.

• Fourier series and Parseval's identity. This is the method that is often used to compute the sum $\sum_{n=1}^\infty \frac{1}{n^2}$. The summand is split into a product of Fourier coefficients of two known functions and the sum is transformed into an integral of the product of these functions. (One can also use a slight variation of the method with power series instead of Fourier series, as shown here.) While the integral in question can be factored in multiple ways that give Fourier series of known functions (e.g., with Mathematica's help). Their product doesn't seem to be of the type that would have a closed form itself.

Since these methods don't seem to be succeeding, it's likely that there is no answer in commonly used closed forms.