Below is a closed-form evaluation in terms of simple functions. Let $y=z/(z-1).$ Then $${}_3F_2(2,n,n;1,n+1;z)=n(-z)^{-n}\Big((n-1)\big(\log(1-y)+\sum_{k=1}^{n-1}\frac{y^k}{k} \big) + y^n \Big)$$ I proved this by using Pochhammer symbol properties, which gets me to a linear combination of ${}_2F_1.$ Then I used a linear transformation to get from the argument $z$ to $y.$ That series can be manipulated to give the logarithm and a finite sum. Variable $y$ is always negative, but if it is small, the sum will converge rapidly. If you put this on a computer, watch out for large negative $y,$ which occurs for $z$ close to 1. In the finite sum, I'd probably add terms pairwise. If the $z\sim 1$ case is your most important case, then it's probably worth thinking about this some more.

Below is a closed-form evaluation in terms of simple functions. Let $y=z/(z-1).$ Then $${}_3F_2(2,n,n;1,n+1;z)=n(-z)^{-n}\Big((n-1)\big(\log(1-y)+\sum_{k=1}^{n-1}\frac{y^k}{k} \big) + y^n \Big)$$ I proved this by using Pochhammer symbol properties, which gets me to a linear combination of ${}_2F_1.$ Then I used a linear transformation to get from the argument $z$ to $y.$ That series can be manipulated to give the logarithm and a finite sum. Variable $y$ is always negative, but if it is small, the sum will converge rapidly. If you put this on a computer, watch out for large negative $y,$ which occurs for $z$ close to 1. In the finite sum, I'd probably add terms pairwise. If the $z\sim 1$ case is your most important case, then it's probably worth thinking about this some more.

Added: The sum for $z\sim 1$ can be done by an expansion found in 'Asymptotic Expansions Pertaining to the Logarithmic Series and Related Trigonometric Sums,' G. Fikioris & P. Andrianesis, J. Class. Analysis vol 7 #2, (2015) 113-127. $$ \sum_{k=1}^{n-1}\frac{y^k}{k} \sim y^n\sum_{k=0}^\infty \frac{A_k(y)}{(y-1)^{k+1}} \frac{1}{n^{k+1}} \, ,n \to \infty $$ where the $A_k(y)$ are the Eulerian polynomials and begin with $A_0(y)=1$ and $A_1(y) = y.$