The answer is yes, this always holds.

Note that

$$\dim(im(f)) \cdot \|f\|^2  \cdot | {\rm supp}(f)| \geq \tau(f^*f) \cdot |{\rm supp}(f)| \geq |G|  \cdot \|f\|^2_1.$$

Here, $\tau \colon \mathbb C[G] \to \mathbb C$ is the non-normalized trace on $\mathbb C[G]$, coming from the inclusion $\mathbb C[G] \subset M_{|G|} \mathbb C$. It is best decribed by $\tau(\sum_{g \in G} a_g g) = |G| \cdot a_e$. The first inequality, follows since $\tau(f^*f)$ is just the sum of the eigenvalues of $f^*f$, which is obviously bounded by the number of non-zero eigenvalues (which equals the dimension of image of $f$) times the size of the largest eigenvalue (which equals $\|f\|^2$). Note that by direct computation
$$\tau\left((\sum_{g \in G} a_g g)^* (\sum_{g \in G} a_g g)\right)=|G| \cdot \sum_{g \in G} |a_g|^2.$$ The second inequality follows from this observation and the Cauchy-Schwarz inequality applied to $f\cdot \chi_{{\rm supp} f}$, where the product is here the pointwise product of coefficients and $\|f\|_1$ denotes the usual 1-norm on $\mathbb C[G]$.

Now, since each group element acts as a unitary (and hence with operator norm $1$) on $M_{|G|} \mathbb C$, we get $\|f\|_1 \geq \|f\|$ and hence

$$\dim(im(f)) \cdot | {\rm supp}(f)| \geq |G|.$$

This even has an extension to *all* (possibly infinite) groups with essentially the same proof. The appropriate statement is then that for the normalized Murray-von Neumann dimension (with respect to the group von Neumann algebra $LG$) of the closure of the image of $\lambda(f)$ acting on $\ell^2 G$ via the left-regular representation $\lambda$, we have

$$\dim_G \left (\overline{im(\lambda(f))} \right) \cdot |{\rm supp}(f)| \geq 1.$$