Noether's (first) theorem is not applicable to local symmetries, such as gauge symmetries. For local symmetries, one should instead use the less-well-known second theorem of Noether. In general, one can state both her first and second theorem as follows:
Let $\mathscr{L}(\phi):= \frac{\partial L}{\partial \phi} - \partial_\mu \frac{\partial L}{\partial (\partial_\mu \phi)}$ denote the Lagrange expression for the field $\phi$. Let $\phi_i$ be the fields of our theory. Then, for any variation of the fields ($\delta\phi_i = \delta_0 \phi_i + \partial_\mu\phi_i\delta x^\mu$) and coordinates $$ \sum_i \mathscr{L}(\phi_i)\delta_0 \phi + \sum_i \partial_\mu (L\delta x^\mu + \frac{\partial L}{\partial(\partial_\mu \phi_i)}\delta_0\phi_i) = 0$$ holds if $\delta S = 0$.
From this, Noether's first theorem with the usual conserved current follows by taking the variation of the fields to be a global symmetry transformation, i.e. $\delta \phi_i = \sum_a \frac{\partial \delta \phi_i}{\partial \epsilon^a}\epsilon^a$ for the continuous parameter $\epsilon \in \mathbb{R}^N$ of the symmetry group, and recognizing that, on-shell, $\mathscr{L}(\phi_i) = 0$.
Noether's second theorem is what you get when the $\epsilon^a$ are allowed to depend on spacetime. If we allow such local transformations, then we have $\delta \phi_i = c_{ai}\epsilon^a + d^\mu_{ai}\partial_\mu\epsilon^a$ for some constants $c,d$. (In general, the local transformation may be allowed to depend on higher derivatives, too, but in the context of gauge theories it won't (and even there, the only derivative-dependent transformation is that of the gauge field itself)) Then, the statement of Noether's second theorem is
$$ \sum_i \mathscr{L}(\phi_i)c_{ai} - \sum_i \partial_\mu (\mathscr{L}(\phi_i)d^\mu_{ai}) = 0$$
In the case of electromagnetism, you have (at least) two ways to use these variations on Noether's theorem to get charge conservation: You can use the global gauge symmetry together with the matter field Euler-Lagrange equations, or you can use the local gauge symmetry together with the gauge field E-L equations (you have to be a bit tricky to do it the latter way - see the reference at the end).
The naive conserved current from the question, $J^\mu = F^{\mu\nu}\partial_\nu \chi$, has no physical significance, since it is not gauge-invariant, and hence not observable.
A wonderfully concise tour of the different versions of Noether's theorem in the context of gauge theories is given in "Noether's Theorems and Gauge Symmetries" by K. Brading and H. R. Brown.
May I ask what text you are reading? My understanding of the stress energy tensor is as follows. The Noether condition is written as,\begin{equation}
\partial _\mu \bigg[\frac{\partial \mathcal L}{\partial (\partial _\mu \phi )}\delta \phi +\mathcal L \delta x^\mu\bigg]=0
\end{equation}
In the discrete case we can imagine separate infinitesimal time and space translations. Field theory kind of smooshes (technical term) space and time together into a spacetime manifold. If we consider an active infinitesimal transformation $x^\nu \rightarrow x^\nu -\lambda ^\nu$, hence $\phi(x)\rightarrow \phi(x+\lambda)=\phi(x)+\lambda^\nu \partial _{\nu}\phi(x)$. If the Lagrange density is not an explicit function $x^\nu$ then we expect the following to be conserved,
\begin{equation}
\frac{\partial \mathcal L}{\partial (\partial _\mu \phi)}\partial _\nu \phi -\delta ^\mu_\nu \mathcal L\nonumber
\end{equation}
In such a transformation the form variation $\delta \phi$ is only dependent of the derivatives of the field so,
\begin{equation}
\delta \phi \longrightarrow \lambda ^\nu \partial _\nu \phi
\end{equation}
The 4-vector variation is $-\lambda^\nu$,
\begin{equation}
\delta x^\mu \longrightarrow -\lambda ^\nu
\end{equation}
Pull out the $\lambda ^\nu$. Then $\partial _{\mu}$ is non-zero if $\delta ^{\mu}_{\nu}$.
Please let me know if you don't agree!
Best Answer
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.