There exists a norm on $Y$ making $L$ bounded if and only if the kernel of $L$ is closed.

However, $\|y\|_Y = \inf\lbrace \|x\|: L(x)=y\rbrace$ is always a *semi-norm* (that is
$\|y\|_Y$ may be $0$ for $y\neq 0$) which induces the final topology $\tau_L=\lbrace A\subseteq Y: L^{-1}(A)$ is open in $X\rbrace$. This is simple: $\tau_L$ is finer than
the $\|\cdot\|_Y$-topology since $L$ is continuous w.r.t. $\|\cdot\|_Y$. Conversely, if
$y\in A \in \tau_L$ you find $x\in X$ with $L(x)=y$ and since $L^{-1}(A)$ is open there
is $\varepsilon>0$ with $T(B(x,\varepsilon)) \subseteq A$ which implies that $A$
contains $B(y,\varepsilon)$.