You're right, they both look similar at first glance. You could interpret the sample variance of a random variable as the average sum of squared errors, where the error is the deviation of a sample from the sample mean of the given random variable. We have to be careful not to confuse an estimate of the variance with the true population variance though.

Given a true population mean $\mu$, aswell as all $N$ data samples from your distribution, we can calculate the variance as the average expected deviation of the samples from the population mean. 

$$
Var(X) = \frac{1}{N}\sum_{i=1}^N (x_i - \mu)^2
$$

Whereas, if we interpreted the error as the deviation from the mean, the sum of squared errors would just be

$$
\sum_{i=1}^N (x_i - \mu)^2
$$

You can see, that the sum of squared errors is missing the correction factor $\frac{1}{N}$, in order to be equal to the population variance.
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