Carbon dating: What to compare the new value against?
Large elements of my answer is drawn from my own notes that I use for teaching General Chemistry II.



What is −0.693?

Radioactive decay processes follow first-order kinetics.  A first-order reaction is one where the rate depends only on the concentration of one of the reactants raised to the first power.  So consider the following reaction:

$$\mathrm{A} \longrightarrow \mathrm{products}$$

The rate can be expressed as the rate of change in the reactant concentration with a change in time such that

$$\mathrm{rate} = -\dfrac{\Delta[\mathrm{A}]}{\Delta t}$$

This simply means that [A] is being consumed (i.e. converted to products) as the reaction proceeds with time.  The corresponding rate law can be written as

$$\mathrm{rate} = k[\mathrm{A}]$$

where $k$ is a rate constant.  We can set these two expressions equal to each other to give

$$-\dfrac{\Delta[\mathrm{A}]}{\Delta t} = k[\mathrm{A}]$$

At this point we must apply a bit of calculus.  Let us write the previous equation in differential form to give

$$-\dfrac{d[\mathrm{A}]}{dt} = k[\mathrm{A}]~~~\mathrm{or} 
~~~\dfrac{d[\mathrm{A}]}{[\mathrm{A}]} = -kdt$$

Integrate over $t=0$ to $t = t$ to give

$$\int_{[\mathrm{A}]_0}^{[\mathrm{A}]_t}\dfrac{d[\mathrm{A}]}{[\mathrm{A}]} = -k\int_0^t dt$$

which results in

$$\ln[\mathrm{A}]_t - \ln[\mathrm{A}]_0 = -kt$$

or

$$ \ln\dfrac{[\mathrm{A}]_t}{[\mathrm{A}]_0}=-kt$$

where $[\mathrm{A}]_0$ is an initial concentration and $[\mathrm{A}]_t$ is the concentration at some time, $t$.  

Now that we have our first-order integrated rate law, we can now discuss half-life.  Half-life (denoted as $t_{\frac{1}{2}}$) is simply the time required for some reactant concentration to drop to half its original value (where the original value is simply [A]$_0$.  So

$$t = t_{\frac{1}{2}}~~~\mathrm{when}~~~[\mathrm{A}]_t = \dfrac{1}{2}[\mathrm{A}]_0$$

Given this relationship, we can rearrange our first-order integrated rate law to solve for $t$ such that

$$\ln\dfrac{[\mathrm{A}]_t}{[\mathrm{A}]_0} = -kt~~~\longrightarrow~~~
t = \dfrac{1}{k}\ln\dfrac{[\mathrm{A}]_0}{[\mathrm{A}]_t} $$

Therefore, we can make some simple substitutions into our first-order integrated rate law at $t = t_{\frac{1}{2}}$ to give

$$t_{\frac{1}{2}} = \dfrac{1}{k}\ln\dfrac{[\mathrm{A}]_0}{\frac{1}{2}[\mathrm{A}]_0} $$

We can then simplify this equation such that

$$t_{\frac{1}{2}} = \dfrac{1}{k}\ln2\dfrac{[\mathrm{A}]_0}{[\mathrm{A}]_0} ~~~\longrightarrow~~~
t_{\frac{1}{2}} = \dfrac{1}{k}\ln (2) 
~~~\longrightarrow~~~
t_{\frac{1}{2}} = \dfrac{0.693}{k}$$

So to answer your question, the 0.693 value comes from the $\ln(2)$ term.  As ringo pointed out below (and I will include here to be thorough), you can now solve your first-order integrated rate law for $t$ and substitute in your half-life equation such that

$$\ln\dfrac{[\mathrm{A}]_t}{[\mathrm{A}]_0}=-kt
~~~\longrightarrow~~~
\dfrac{\ln\frac{[\mathrm{A}]_t}{[\mathrm{A}]_0}}{-k}=t
~~~\longrightarrow~~~
\dfrac{\ln\frac{[\mathrm{A}]_t}{[\mathrm{A}]_0}}
{
-\frac{0.693}{t_{\frac{1}{2}}}
}
=t
$$

or more cleanly...

$$t = \dfrac{\left(t_{\frac{1}{2}}\right)
\left(\ln\frac{[\mathrm{A}]_t}{[\mathrm{A}]_0}\right)}
{-0.693}
$$

How do you get $N_0$?

You could get $N_0$ by examining the activity of carbon-14 in a relatively recent object or plant.  Carbon-14 in a living plant is not expected to have decayed at all (since the half-life is over 5000 years!).  Therefore you could use its activity for $N_0$.  Of course this procedure relies on a few assumptions but carbon-14 dating is an approximation to begin with.