One way you can combat this is to use proportions in each category, which does not require you to put numbers in for each category (you can leave it as 80% rated as "strongly likes"). However proportions do suffer from the small number of ratings issue. This shows up in your example the Photo with 1 +5 rating would get a higher average score (and proportion) than one with the 99 +5 and 1 +2 rating. This doesn't fit well with my intuition (and I suspect most peoples).
One way to get around this small sample size issue is to use a Bayesian technique known as "Laplace's rule of succession" (searching this term may be useful). It simply involves adding 1 "observation" to each category before calculating the probabilities. If you wanted to take an average for a numerical value, I would suggest a weighted average where the weights are the probabilities calculated by the rule of succession.
For the mathematical form, let $n_{sd},n_{d},n_{l},n_{sl}$ denote the number of responses of "strongly dislike", "dislike", "like", and "strongly like" respectively (in the two examples, $n_{sl}=1,n_{sd}=n_{d}=n{l}=0$ and $n_{sl}=99,n_{l}=1,n_{sd}=n_{d}=0$). You then calculate the probability (or weight) for strongly like as
$$Pr(\text{"Strongly Like"}) = \frac{n_{sl}+1}{n_{sd}+n_{d}+n_{l}+n_{sl}+4}$$
For the two examples you give, they give probabilities of "strongly like" as
$\frac{1+1}{1+0+0+0+4}=\frac{2}{5}$ and $\frac{99+1}{99+1+0+0+4}=\frac{100}{104}$ which I think agree more closely with "common sense". Removing the added constants give $\frac{1}{1}$ and $\frac{99}{100}$ which makes the first outcome seem higher than it should be (at least to me anyway).
The respective scores are just given by the weighted average, which I have written below as:
$$Score=\begin{array}{1 1} 5\frac{n_{sl}+1}{n_{sd}+n_{d}+n_{l}+n_{sl}+4}+2\frac{n_{l}+1}{n_{sd}+n_{d}+n_{l}+n_{sl}+4} \\ - 2\frac{n_{d}+1}{n_{sd}+n_{d}+n_{l}+n_{sl}+4}
-5\frac{n_{sd}+1}{n_{sd}+n_{d}+n_{l}+n_{sl}+4}\end{array}$$
Or more succinctly as
$$Score=\frac{5 n_{sl}+ 2 n_{l} - 2 n_{d} - 5 n_{sd}}{n_{sd}+n_{d}+n_{l}+n_{sl}+4}$$
Which gives scores in the two examples of $\frac{5}{5}=1$ and $\frac{497}{104}\sim 4.8$. I think this shows an appropriate difference between the two cases.
This may have been a bit "mathsy" so let me know if you need more explanation.
A simple version of ELO can be cast as a logistic regression: for players $i,j$ with ratings $R_i,R_j$,
$$P(i\mbox{ beats }j)=\frac{1}{1+\exp(\beta(R_i-R_j))}.$$
So you could just as easily predict score instead by using a different link function, for example a lorentzian or gaussian:
$$P(\mbox{Game score}=x)=a\exp(-\alpha|\beta(R_i-R_j)-x|^\gamma)$$,
where the game score can be positive (in favor of $i$) or negative (in favor of $j$). So you don't need to calculate the probability of beating and just directly optimize the game score.
Best Answer
Evan Miller's how not to sort by average rating has pretty good advice for the general problem you are trying to solve.