The Coulomb potential has the following forms for a positive charge in each dimensionality:
\begin{align}
\Phi_{\operatorname{1-d}}(r) &= -\frac{\sigma}{2\epsilon_0} r, \\
\Phi_{\operatorname{2-d}}(r) &= -\frac{\lambda}{2\pi \epsilon_0} \ln(r),\ \mathrm{and} \\
\Phi_{\operatorname{3-d}}(r) &= \frac{q}{4\pi\epsilon_0} \left(\frac{1}{r}\right).
\end{align}
The reason for this is that the electric field, defined as $-\nabla\Phi$ in general and $-\dfrac{\partial\Phi}{\partial r} \hat{r}$ in this case, times the measure of the boundary of a "ball" has to be a constant. In 1-d, a ball is a line and the measure of its boundary is just the number of points on its ends (i.e. 2). In 2-d the ball is a circle and the measure of its boundary is the circumference (i.e. $2\pi r$). In 3-d the ball is a sphere and the measure of its boundary is the sphere's surface area ($4\pi r^2$). Notice that those are exactly the quantities in the denominator when we calculate the electric field from the potentials:
\begin{align}
\vec{E}_{\operatorname{1-d}}(r) &= \frac{\sigma}{\epsilon_0} \left(\frac{\hat{r}}{2}\right), \\
\vec{E}_{\operatorname{2-d}}(r) &= \frac{\lambda}{\epsilon_0} \left(\frac{\hat{r}}{2\pi r}\right),\ \mathrm{and} \\
\vec{E}_{\operatorname{3-d}}(r) &= \frac{q}{\epsilon_0} \left(\frac{\hat{r}}{4\pi r^2}\right).
\end{align}
The name of the law that implies this is known as Gauss's law.
The objects which are designed to hold the most charge for the least mass are the capacitors which you can buy very cheap from any electronics store, such as RS Components. They do this by storing an equal amount of +ve and -ve charge in close proximity yet separated by an insulator with a high dielectric constant, which increases the charge-storing storage capacity by orders of magnitude compared with having air between the stored charges. Wires will not be anywhere near as effective.
The specification of particular capacitors will tell you the maximum voltage $V$ which can be applied. This is related to charge stored $Q$ and capacitance $C$ by $Q=CV$.
Conducting wire can be modelled as a coaxial capacitor with the outer plate at infinity. In practice the outer plate will be any of several nearby conducting surfaces, which could be anything between a few mm to a few m away.
The capacitance of a wire of length $1m$ and diameter $0.1mm$ is about $10pF$ (pF=$10^{-12}$ Farads). Typical electrolytic capacitors are between $0.1\mu F$ and $0.1mF$, a factor of $10^4$ to $10^7$ times larger - ie for the same applied voltage they can store around a million times as much charge.
Charge storage capacity is limited by the breakdown strength of the insulating material between the capacitor plates. In the case of bare wire this is air, which breaks down at a field strength of $E=3kV/mm$. (Mica, used in old capacitors, breaks down at $118kV/mm$.) Using the relation $E=\frac{\sigma}{\epsilon_0}$ this requires a surface charge density $\sigma$ of about $30pC/mm^2$. Your $1m$ wire would have an area of about $600mm^2$. Assuming a uniform charge density, it could hold at most $20nC$. With capacitance of $10pF$ the breakdown voltage would be $V=Q/C=20nC/10pF=2kV$.
The weak points for wire, at which air will break down will occur, are the sharp ends, where surface charge density and electric field are highest. So with a more rigorous calculation the maximum voltage and charge stored will be significantly lower.
Best Answer
In principle, you can charge a conductor indefinitely.
But remember that in order to cause a flow of charges from a body (call it a 'source') to another (the conductor in question), the potential of the former has to be lower than the potential of the latter.
This potential difference causes a current to flow from the source to the conductor, resulting in a transfer of charges. All the charges that are already in the conductor will exert an electrostatic repulsion on the incoming ones, slowing them down and making it more and more difficult (as more and more charge accumulates in the conductor) to charge it further. As charge continues to flow out of the source and into the conductor, the potential difference decreases and the potentials of the two objects become more and more similar.
At some point, the potential difference will reach 0. The current will stop flowing when the charge in the source is equal to the charge in the conductor, which corresponds to the situation in which the electrostatic repulsion from the charges in the conductor is equal to the force attempting to put more charges in. Here, the potential difference is 0.
If you want to force more charges to flow out of the source and into the conductor, therefore increasing its charge, you have two options:
1) you either increase the potential of the source, so as to create a non-zero potential difference and thus causing current to flow.
2) introduce a new force that pushes the charges out of the source and into the conductor: this force (eg chemical force) has the job of bringing a charge of the source against the electric field exerted by the charges of the conductor.
The real limit to the charging of a conductor would be when there is no physical space available for the electrons. Electrons are fermions and they cannot occupy the same position in space, so it will become harder and harder to squash them together.
NOW, in the context of capacitance:
Imagine a circuit with a generator (battery) and a capacitor. DC current will flow only in half of the circuit (left or right depending on the choice of charge carries, electrons or ions, that is conventional). Let's say protons carry the charge (this is the convention for current, although physically it is the electrons that to the moving).
Protons leave the + terminal and accumulate on the closest plate of the capacitor. The plate is now the conductor, and the + terminal of the battery is the source. The plate has initially 0 charge and therefore 0 potential. The charges accumulating on the plate will exert an electric field that is going to oppose incoming protons. The potential of the + terminal decreases, the potential of the plate increases, the potential difference reaches 0 and current stops flowing: there are as many protons in the + terminal as there are on the plate.
Unless you increase the charge in the + terminal (therefore increasing its potential) or apply another force on the protons, no more current will flow.
If you now disconnect the battery and close the circuit, you have one of the capacitor plate charged (so at a non zero potential) and the other one with no charge ( 0 potential). Potential difference => Current will flow until the charge on both plates is the same.