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How can one calculate the order of a linear group with elements in Z/nZ? I imagine it has something to do with factorials but I'm not sure of the actual calculations used.

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Let's do $GL(n, \mathbb Z_p)$ first. How many matrices are there in this group? Let's count the possibilities row by row. For the first row, anything is allowed except $(0, \dots, 0)$. For the second row, anything is allowed that isn't a multiple of the first row. For the third row, anything is allowed that is linearly independent of the first two rows etc. Does this make sense to you?

For $SL(n, \mathbb Z_p)$, consider the group homomorphism $GL(n, \mathbb Z_p) \to \mathbb Z_p^{\times}$ given by taking determinants. Its image is the whole of $Z_p^{\times}$. Its kernel is $SL(n, \mathbb Z_p)$. Does this help?

Finally, $PSL(n, \mathbb Z_p)$ is just $SL(n, \mathbb Z_p)$ quotiented by the matrices of the form "constant times the identity". This should be easy once you have done the previous parts.

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    This may help: https://proofwiki.org/wiki/Order_of_General_Linear_Group_over_Finite_Field – 2017-02-28
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    "Its image is the whole of..." should end with $(\mathbb{Z}/p\mathbb{Z})^\times$. And I guess you mean $\det: \mathrm{GL}_n(\mathbb{Z}/p\mathbb{Z}) \to (\mathbb{Z}/p\mathbb{Z})^\times$, too. – 2017-02-28
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    Yes, well spotted! – 2017-02-28