When you work with variances, know these facts (in addition to the definition of variance):
The variance of a sum of independent (or just uncorrelated) variables is the sum of their variances.
The expectation of a sum of variables (independent or not) is the sum of their expectations.
The expectation (of any discrete variable) is the sum of each possible value multiplied by its probability.
As an example, let's apply them to your case. You may model the sum of your Bernoulli trials with a variable $X$ expressed as the sum of $n$ independent variables $X_i$. Each $X_i$ takes on the value $1$ with probability $p$ and the value $-1$ with the probability $1-p$.
Fact $(3)$ asserts $$\mathbb{E}(X_i) = p(1) + (1-p)(-1) = 2p-1$$ and $$\mathbb{E}(X_i^2) = p(1)^2 + (1-p)(-1)^2 = p + (1-p) = 1.$$
Fact $(2)$ asserts $$\eqalign{\mathbb{E}(X) &= \mathbb{E}(X_1+\cdots+X_n) = \mathbb{E}(X_1) + \cdots + \mathbb{E}(X_n) = n\mathbb{E}(X_1) \\&=n(2p-1).}$$
The definition of variance now tells you$$\operatorname{Var}(X_i) = \mathbb{E}(X_i^2) - \mathbb{E}(X_i)^2 = 1 - (2p-1)^2 = 4p(1-p).$$
Consequently, fact $(1)$ yields $$\eqalign{\operatorname{Var}(X) &= \operatorname{Var}(X_1+\cdots+X_n) = \operatorname{Var}(X_1) + \cdots + \operatorname{Var}(X_n)=n \operatorname{Var}(X_1) \\&= n(4p(1-p)).}$$
The appearance of the factor of $4$ might seem somewhat mysterious. There's another extremely useful fact you might consider using to shortcut these considerations and identify the origin of that factor:
(4) The variance of a shifted, rescaled variable is the square of the scale factor times the variance. In mathematical symbols, $$\operatorname{Var}(\sigma X + \mu) = \sigma^2 \operatorname{Var}(X)$$ no matter what values the numbers $\sigma$ and $\mu$ might have.
This applies by noting that your $X_i$ can all be expressed as $2Y_i-1$ where $Y_i$ is a true Bernoulli (that is, $0-1$) variable. It's simple to show this: since $2\times 1-1=1$ and $2\times 0 - 1=-1$, the values of $1$ and $0$ taken on by $Y_i$ become $1$ and $-1$, respectively, for $X_i$. The probabilities are unchanged.
You might already know that $Y=Y_1+\cdots + Y_n$, the sum of $n$ independent Bernoulli variables with common probability $p$, is called a Binomial variable. It has a Binomial distribution. You can remember or look up its variance, which is $np(1-p)$. Since $$X = X_1+\cdots+X_n = (2Y_1-1) + \cdots + (2Y_n-1) = 2(Y_1 + \cdots + Y_n) - n=2Y-n,$$ you can take $\sigma=2$ in applying fact $(4)$, which immediately tells you $$\operatorname{Var}(X) = 2^2 \operatorname{Var}(Y) = 4np(1-p).$$