If $p \geq 2$, then an appropriate version of the Hölder inequality shows that for $U$ bounded, $L^p(U) \hookrightarrow L^2(U)$, hence $W^{1,p}\hookrightarrow W^{1,2} \hookrightarrow L^2$ using the trace inequality. So you have
$ \|u\|_2 \leq \|u\|_2^{1-1/p}\|u\|_2^{1/p} \lesssim \|u\|_p^{1-1/p} \|u\|^{1/p}_{W^{1,2}} \lesssim \|u\|^{1-1/p}_p \|u\|^{1/p}_{W^{1,p}} $
which does not require passing through the Gagliardo-Nirenberg inequality. (Though one can argue that the proof of the GN inequality can be recycled to prove the trace theorem, so the two aren't completely unrelated.)
The interesting case is when $p < 2$, in which your estimate follows partially from the generalized Sobolev imbedding theorems (which allows also for trace estimates). See, for example, Theorem 4.12 Case C of Robert Adams' Sobolev Spaces. In particular the numerology requires $p \leq 2 \leq (n-1)p / (n-p)$ which implies
$ \frac{2n}{n+1} \leq p \leq 2 $
must hold for the classical trace theorem $W^{1,p}(\Omega) \hookrightarrow L^2(\partial\Omega)$. (If you believe in fractional Sobolev spaces, then you can also "obtain" the above by combining a fractional version of Gagliardo-Niremberg from $W^{1,p}\hookrightarrow H^s$ for some $s > 1/2$ and using the fractional trace theorem $H^s(\Omega)\hookrightarrow L^2(\partial\Omega)$; note that this route naively will require the lower bound on $p$ to be a strict inequality due to the failure of $H^{1/2}(\Omega)\not\hookrightarrow L^2(\partial\Omega)$.) In particular, for dimension $n > 1$ the $L^1$ endpoint is not covered by this theorem.
This failure at $L^1$ is not a problem of the method: it is a genuine failure of the inequality. Consider the function in two dimension $u(x,y) = \left[ (x-1)^2 + y^2\right]^{1/4}$, which is just a translated version of the function $\frac{1}{\sqrt{r}}$. It is not $L^2$ integrable when restricted to the circle, since it has a Logarithmic singularity at $(1,0)$. But $\frac{1}{\sqrt{r}}$ is certainly $L^1_{loc}(\mathbb{R}^2)$, and its derivative which has a $\frac{1}{r^{3/2}}$ singularity is also $L^1_{loc}(\mathbb{R}^2)$. (The same example also works in arbitrary $n \geq 2$; you can either use the same principle via functions like translates of $\frac{1}{r^{(n-1)/2}}$, or just argue through the method of descent.)