This must be a misunderstanding as to how RAID 5 or other schemes work.
With N disks each of size M or larger, the storage capacity is (N-1)*M for a RAID 5 layout. The parity information is distributed all over the members of the disk set. For example, with 4 (N) disks each of size 4 TiB (M), you get an RAID 5 volume of 12 TiB.
The least "efficient" RAID scheme is RAID 1/10, which in the example above would provide N*M/2 or an 8 TiB volume. RAID 6 results in (N-2)*M thus in this example also 8 TiB. There are numerous other RAID schemes but these are hardly practical for the comparatively small disk arrays of the home user.
The storage used for the operating system of the NAS is very small so can be safely ignored in these calculations.