Law of Conservation of Energy on chain falling from table

Your error is in ignoring the potential energy of the mass on the table. You can ignore it only if you take the point of reference as the table, which you haven't. Also, in your choice of reference, $h$ is not given in the question, so it is unclear where exactly your reference is.

You can solve this sum by consider the sections of the chain as particles located at their centres of mass. Considering table as reference, you will get initial potential energy as $-frac {mgL}{32}$, and final potential energy as $-frac {mgL}{2}$. You can then apply conservation of energy to get kinetic energy and the velocity of the chain.

Consider instead a force balance.

Let $x$ represent the downward axis with origin at the table edge. Also, let $l_0$ be the initial length over the edge of the table. We assume no friction here, and momentum is conserved.

$F = frac{x}{L}mg$ and applying Newton's 2nd law, $mddot{x} = frac{x}{L}mg$. This yields a simple, first order ODE, and upon applying the initial conditions $x(0)=l_0$ and $dot{x}(0) = 0$, then letting $k=sqrt{g/L}$,

$x(t) = frac{l_0}{2}left(e^{kt} + e^{-kt} right) = l_0 cosh(kt)$

The moment the last part of the chain leaves the table, time $t^*$, we have $L = l_0 cosh(kt^*)$ or $t^* = frac{1}{k}text{arcosh}(L/l_0)$

Then, with $x=L/l_0$, and some identities

begin{align} v^* &= l_0 k sinh(text{arcosh}(x)) &= l_0 k sinh(ln (x + sqrt{x^2 - 1})) &= frac{l_0 k }{2}left(x + sqrt{x^2 - 1} - frac{1}{x + sqrt{x^2 - 1}}right) &= frac{l_0 k }{2}left( 2sqrt{x^2 - 1}right) end{align}

Substituting in the values and simplifying yields the final answer,

$v^* = sqrt{frac{15 L g}{16}} $