What is the shape of [Cr(NO)₄]?
Based on available theoretical considerations and a available literature, a tetrahedral geometry appears to be a good assumption for $\ce{Cr(NO)4}$.

The NO ligand can be rendered as a three-electron donor, if we take
it as uncharged.  It is actually similar to CO, really, in that it
interacts through its $\pi^*$ orbitals as well as its $\sigma$
orbital; but NO unlike CO has an electron to offer from its $\pi^*$.
That plus the $\sigma$ electrons makes three.
We then may identify the central metal atom as having 18 valence
electrons.  When this happens with multiple ligands we typically see
an arrangement with highest order symmetry, thus tetrahedral for
four-coordination.  This gives the best possible bonding to the
entire set of valence orbitals involved in the 18-electron
structure.  This is seen in the tetrahedral four-coordinationn of
$\ce{Ni(CO)4}$, and in the octahedral six-coordination of the series
$\ce{[V(CO)6]-, Cr(CO)6, [Mn(CO)6]+, [Fe(CO)6]^{2+}}$.  So, we should
expect $\ce{Cr(NO)4}$ to be tetrahedral.
Heberhold and
Razavi
found evidence for a tetrahedral structure in $\ce{Cr(NO)4}$ from
its IR spectrum.