How is the energy level of an atom's valence electrons related to its period in the periodic table?

There is no strict relationship, but for NON-transition metals (i.e. non-d-block, non-f-block), there is one.

Valence electrons are then USUALLY listed after the noble gas core, within reason (e.g. tungsten, #"W"#, probably doesn't have #20# valence electrons, but up to #6# instead).

For NON-transition metals, the energy level #n# is given by its period number, i.e. its row number. Being on the fourth row of the periodic table means your valence electrons would be the #4s# and #4p# electrons, #n = 4#.

Two examples:

#"Ca"#, with configuration #[Ar]color(blue)(4s^2)#. This has #bb2# valence electrons.

#"P"#, with configuration #[Ne]color(blue)(3s^2 3p^3)#. This has #bb5# valence electrons.

EXCEPTIONS BELOW!

Transition metals on the other hand have easy access to occupied #(n-1)d# orbitals, and thus have the electrons in those orbitals included in their set of valence electrons.

Three examples:

#"Sc"# (scandium), with configuration #[Ar] color(blue)(3d^1 4s^2)#. This is why #"Sc"# can have a maximum oxidation state of #+3# (e.g. in #"ScCl"_3#); it has #bb3# valence electrons.

#"W"# (tungsten), with configuration #[Xe]color(red)(4f^14) color(blue)(5d^4 6s^2)# --- note that the #4f# electrons are hardly used, even though they are listed after "#[Xe]#", the noble gas core.
#"W"# is commonly going to have a #+6# maximum oxidation state (e.g. in #"WO"_3#), which means it probably has #bb6# valence electrons most of the time.

#"Os"# (osmium), with configuration #[Xe]color(red)(4f^14) color(blue)(5d^6 6s^2)#. It has up to #bb8# valence electrons, e.g. in #"OsO"_4#.

Some heavy #bb(f)#-block metals (mainly lanthanides and actinides) also have access to occupied #(n-2)f# orbitals too, and those electrons might also be included in their set of valence electrons... they might even not have #(n-1)d# valence electrons sometimes.

Three examples:

#"Pa"# (protactinium), with configuration #[Rn]color(blue)(5f^2 6d^1 7s^2)#. This is why #"Pa"# usually has a #+5# oxidation state in its compounds (such as #"Pa"_2"O"_5#); it has #bb5# valence electrons.

#"Bk"# (berkelium), with configuration #[Rn]color(red)(5f^9) color(blue)(7s^2)# --- note that #11# valence electrons would be insane. In berkelium, only some of the #5f# electrons are considered valence, depending on context.
Since #+4# is the highest known easily-accessible oxidation state (e.g. in #"BkO"_2#), #"Bk"# has around #bb4# valence electrons (not #2#!).

#"Gd"# (gadolinium), with configuration # [Xe] color(red)(4f^7) color(blue)(5d^1 6s^2)#. It most reasonably has #bb3# valence electrons, and indeed, its highest oxidation state is usually #+3# (such as in #"Gd"_2"O"_3#).