Indeed, nothing is wrong with Noether theorem here, $J^\mu = F^{\mu \nu} \partial_\nu \Lambda$ is a conserved current for every choice of the smooth scalar function $\Lambda$. It can be proved by direct inspection, since
$$\partial_\mu J^\mu = \partial_\mu (F^{\mu \nu} \partial_\nu \Lambda)=
(\partial_\mu F^{\mu \nu}) \partial_\nu \Lambda+ F^{\mu \nu} \partial_\mu\partial_\nu \Lambda = 0 + 0 =0\:.$$
Above, $\partial_\mu F^{\mu \nu}=0 $ due to field equations and $F^{\mu \nu} \partial_\mu\partial_\nu \Lambda=0$ because $F^{\mu \nu}=-F^{\nu \mu}$ whereas $\partial_\mu\partial_\nu \Lambda =\partial_\nu\partial_\mu \Lambda$.
ADDENDUM. I show here that $J^\mu$ arises from the standard Noether theorem. The relevant symmetry transformation, for every fixed $\Lambda$, is
$$A_\mu \to A'_\mu = A_\mu + \epsilon \partial_\mu \Lambda\:.$$ One immediately sees that
$$\int_\Omega {\cal L}(A', \partial A') d^4x = \int_\Omega {\cal L}(A, \partial A) d^4x\tag{0}$$
since even ${\cal L}$ is invariant. Hence,
$$\frac{d}{d\epsilon}|_{\epsilon=0} \int_\Omega {\cal L}(A, \partial A) d^4x=0\:.\tag{1}$$
Swapping the symbol of derivative and that of integral (assuming $\Omega$ bounded) and exploiting Euler-Lagrange equations, (1) can be re-written as:
$$\int_\Omega \partial_\nu \left(\frac{\partial {\cal L}}{\partial \partial_\nu A_\mu} \partial_\mu \Lambda\right) \: d^4 x =0\:.\tag{2}$$
Since the integrand is continuous and $\Omega$ arbitrary, (2) is equivalent to
$$\partial_\nu \left(\frac{\partial {\cal L}}{\partial \partial_\nu A_\mu} \partial_\mu \Lambda\right) =0\:,$$
which is the identity discussed by the OP (I omit a constant factor):
$$\partial_\mu (F^{\mu \nu} \partial_\nu \Lambda)=0\:.$$
ADDENDUM2. The charge associated to any of these currents is related with the electrical flux at spatial infinity. Indeed one has:
$$Q = \int_{t=t_0} J^0 d^3x = \int_{t=t_0} \sum_{i=1}^3 F^{0i}\partial_i \Lambda d^3x = \int_{t=t_0} \partial_i\sum_{i=1}^3 F^{0i} \Lambda d^3x -
\int_{t=t_0} ( \sum_{i=1}^3 \partial_i F^{0i}) \Lambda d^3x \:.$$
As $\sum_{i=1}^3 \partial_i F^{0i} = -\partial_\mu F^{\mu 0}=0$, the last integral does not give any contribution and we have
$$Q = \int_{t=t_0} \partial_i\left(\Lambda \sum_{i=1}^3 F^{0i} \right) d^3x = \lim_{R\to +\infty}\oint_{t=t_0, |\vec{x}| =R} \Lambda \vec{E} \cdot \vec{n} \: dS\:.$$
If $\Lambda$ becomes constant in space outside a bounded region $\Omega_0$ and if, for instance, that constant does not vanish, $Q$ is just the flux of $\vec{E}$ at infinity up to a constant factor. In this case $Q$ is the electric charge up to a constant factor (as stressed by ramanujan_dirac in a comment below). In that case, however, $Q=0$ since we are dealing with the free EM field.
Best Answer
Noether's theorem states that if a system has a continuous symmetry, there is a quantity related to this symmetry, called the Noether charge, which is conserved.
It does not state anything on the fact that adding a constant term to a measurable quantity may or may not change the physical description of the system. Only some physical quantities in fact are defined up to a constant term (one can add a constant term without change the physics of the system). These quantities are, for example, some forms of potential energy, angles and angular phases, but not entropy. Entropy is not defined up to a constant term. Adding a constant term to the entropy does change the physics of the system. For example, the 3rd law of thermodynamics states that the entropy of a perfect crystal (or an ideal gas) at zero temperature is zero. This has the very physical consequence that zero temperature can be reached only asymptotically.
Few clarifications of the Noether's theorem
Examples of continuous symmetries: time invariance, translational invariance, rotational invariance. Corresponding conserved charge: energy, linear momentum, angular momentum.
Now, if a system is translational invariant (for example, an isolated and closed system), that means that any thermodynamical observable of the system is translational invariant, e.g., volume, temperature, energy, and entropy.
Note that in a closed system, any irreversible process breaks time invariance in a subtle way, since energy may still be a conserved quantity although entropy is not (it increases). This however does not constitute an exception of Noether's theorem.