Forgive me if I've missed something rather obvious.
I'm a physicist with what is essentially a (histogram) distribution centered about a mean value that approximates to a Normal distribution. The important value to me is the standard deviation of this Gaussian random variable. How would I go about trying to find the error on the sample standard deviation? I have the feeling its something to do with the error on each bin in the original histogram.
It sounds like you're asking for a calculation of the standard deviation of the sample standard deviation. That is, you're asking for ${\rm SD}(s) = \sqrt{ {\rm var}(s) }$, where
$$ s = \sqrt{ \frac{1}{n-1} \sum_{i=1}^{n} (X_i - \overline{X}) }, $$
$X_1, ..., X_n \sim N(\mu, \sigma^2)$ and $\overline{X}$ is the sample mean.
First, we know from the basic properties of variance that
$$ {\rm var}(s) = E(s^2) - E(s)^2 $$
Since the sample variance is unbiased, we know $E(s^2) = \sigma^2$. In Why is sample standard deviation a biased estimator of $\sigma$?, $E(s)$ is calculated, from which we can infer
$$ E(s)^2 = \frac{2 \sigma^2 }{n-1} \cdot \left( \frac{ \Gamma(n/2) }{ \Gamma( \frac{n-1}{2} ) } \right)^2 $$
therefore
$$ {\rm SD}(s) = \sqrt{ E(s^2) - E(s)^2 } = \sigma \sqrt{ 1 - \frac{2}{n-1} \cdot \left( \frac{ \Gamma(n/2) }{ \Gamma( \frac{n-1}{2} ) } \right)^2 } $$
The quantity $X=(n-1) s^2/\sigma^2$ has a chi-squared distribution with $n-1$ degrees of freedom when the samples are independent and distributed with the same normal distribution This quantity can be used to get confidence intervals for the variance of the normal and its standard deviation. If you have the raw values and not just the central value of the bins you can calculate $s^2$.
It is known that if $X$ has a chi-squared distribution with $n-1$ degrees of freedom its variance is $2(n-1)$. Knowing this and the fact the $\mathrm{Var}(cX) = c^2 \mathrm{Var}(X)$ we get that $s^2$ has a variance equal to $$\frac{2(n-1)\sigma^4}{(n-1)^2} =\frac{2\sigma^4}{n-1} \>.$$ Although $\sigma^4$ is unknown you can approximate it by $s^4$ and you have a rough idea of what the variance of $s^2$ is.
There are several ways of quantifying the error of the standard deviation in the normal case. I am going to present the profile likelihood of $\sigma$ which can be used for approximating confidence intervals.
Let $x=(x_1,...,x_n)$ be a sample from a Normal$(\mu,\sigma)$. The corresponding likelihood function is given by
$${\mathcal L}(\mu,\sigma) \propto \dfrac{1}{\sigma^n}\exp\left(-\dfrac{1}{2\sigma^2}\sum_{j=1}^n(x_j-\mu)^2\right)$$
Then, the Maximum Likelihood Estimators are given by $(\hat\mu,\hat\sigma)=(\bar x,s)$, where $s=\sqrt{\dfrac{1}{n}\sum_{j=1}^n(x_j-\bar x)^2}$. Given that you are interested on quantifying the error on $\sigma$, you can then calculate the normalised profile likelihood of this parameter as follows.
$$R_p(\sigma)=\dfrac{\sup_{\mu}{\mathcal L}(\mu,\sigma)}{{\mathcal L}(\hat\mu,\hat\sigma)} = \left(\dfrac{\hat\sigma}{\sigma}\right)^n\exp\left[\dfrac{n}{2}\left(1-\left(\dfrac{\hat\sigma}{\sigma}\right)^2\right)\right]$$
Note that $R_p:{\mathbb R}_+\rightarrow (0,1]$. An interval of level $0.147$ has an approximate confidence of $0.95$. Next I attach an $R$ code that can be used for calculating these intervals. You can modify it accordingly in your context (or if you post the data I can include these changes).
data = rnorm(30)
n = length(data)
sg = sqrt(mean((data-mean(data))^2))
# Profile likelihood
rp = function(sigma) return( (sg/sigma)^n*exp(0.5*n*(1-(sg/sigma)^2)) )
vec = rvec = seq(0.5,1.5,0.01)
for(i in 1:length(rvec)) rvec[i] = rp(vec[i])
plot(vec,rvec,type="l")
rpc = function(sigma) return(rp(sigma)-0.147)
# Approximate 95% confidence interval
c(uniroot(rpc,c(0.7,0.8))$root,uniroot(rpc,c(1.1,1.3))$root)
An advantage of this sort of intervals is that they are invariant under transformations. In this case if you calculate an interval for $\sigma$, $I=(L,U)$, then the corresponding interval for $\sigma^2$ is simply $I^{\prime}=(L^2,U^2)$.
I just want to point out that the standard deviation of $s$ can be well approximated by $\text{SD}(s)\approx \sigma/\sqrt{2n}$ when $n$ is large. The relative error is less than 0.38% for $n=100$ and decreases even further for larger values of $n$.
One way to derive this result applies the asymptotic expansion of the Gamma function to Macro's formula (elsewhere in this thread),
$$\sigma \sqrt{ 1 - \frac{2}{n-1} \cdot \left( \frac{ \Gamma(n/2) }{ \Gamma( \frac{n-1}{2} ) } \right)^2} = \sigma\left(\frac{1}{\sqrt{2n}} + \frac{3}{2^2\sqrt{(2n)^3}} + \frac{15}{2^5\sqrt{(2n)^5}} + \cdots\right)$$
This shows that (at least eventually) the approximation is a little too low. A plot of their ratio shows this is always the case:
