I'm assuming that $n$ is fixed, so that $Y(t)$ is the sum of the previous $n$ observations. You didn't say what the mean and variance of the random variables $X$ was, so for now, I will assume that they are general: $\mu$ and $\sigma^2$. Now your question boils down (by the strong law of large numbers, or by the ergodic theorem) to computing the probability that $Y(t)$ and $Y(t-1)$ have opposite signs. In fact, there are (in the long run) the same number of switches from negative to positive as from positive to negative, so it suffices to compute $\mathbb P(Y(t)>0, Y(t-1)<0)$. Now $Y(t)$ and $Y(t-1)$ are themselves normal random variables with means $n\mu$ and variance $n\sigma^2$. The covariance is $(n-1)\sigma^2$. Now $(Y(t-1),Y(t))$ has a multivariate normal distribution. If $U\sim N(n\mu,(n-\frac 12)\sigma^2)$ and $V\sim N(0,\frac 12\sigma^2)$ are independent normal random variables, then it is easy to see that $(U-V,U+V)$ has the same mean and covariance matrix as $(Y(t-1),Y(t))$. Hence since they are both multivariate normal distributions, they have the same distribution.

Now it suffices to compute $\mathbb P(U-V<0<U+V)$. That is $\mathbb P(V>|U|)=2\mathbb P(V>U>0)$.

At this point, I'll make the assumption that you probably intended $\mu=0$. By scaling the random variabes, you can also assume that $\sigma=\sqrt 2$. Now $V\sim N(0,1)$ and $U\sim N(0,2n-1)$. Let's write $U=\sqrt{2n-1}N'$, so that we are asking for $2\mathbb P(0<N'<V/\sqrt{2n-1})$.

If $n$ is large and the value of $V$ is given and positive, but small in comparison with $\sqrt{2n-1}$ (which it is with overwhelming probability), then the probability that $N'$ is in the correct range is approximately $(1/\sqrt{2\pi})V/\sqrt{2n-1}$.

Hence the desired probability (and hence the frequency of sign changes) is approximately $$\frac{2}{\sqrt {2\pi(2n-1)}}\mathbb E(\max(V,0))=\frac{1}{\pi} (2n-1)^{-1/2}$$.

I'm assuming that $n$ is fixed, so that $Y(t)$ is the sum of the previous $n$ observations. You didn't say what the mean and variance of the random variables $X$ was, so for now, I will assume that they are general: $\mu$ and $\sigma^2$. Now your question boils down (by the strong law of large numbers, or by the ergodic theorem) to computing the probability that $Y(t)$ and $Y(t-1)$ have opposite signs. In fact, there are (in the long run) the same number of switches from negative to positive as from positive to negative, so it suffices to compute $\mathbb P(Y(t)>0, Y(t-1)<0)$. Now $Y(t)$ and $Y(t-1)$ are themselves normal random variables with means $n\mu$ and variance $n\sigma^2$. The covariance is $(n-1)\sigma^2$. Now $(Y(t-1),Y(t))$ has a multivariate normal distribution. If $U\sim N(n\mu,(n-\frac 12)\sigma^2)$ and $V\sim N(0,\frac 12\sigma^2)$ are independent normal random variables, then it is easy to see that $(U-V,U+V)$ has the same mean and covariance matrix as $(Y(t-1),Y(t))$. Hence, since they are both multivariate normal distributions, they have the same distribution.

Now it suffices to compute $\mathbb P(U-V<0<U+V)$. That is $\mathbb P(V>|U|)=2\mathbb P(V>U>0)$.

At this point, I'll make the assumption that you probably intended $\mu=0$. By scaling the random variables, you can also assume that $\sigma=\sqrt 2$. Now $V\sim N(0,1)$ and $U\sim N(0,2n-1)$. Let's write $U=\sqrt{2n-1}N'$, so that we are asking for $2\mathbb P(0<N'<V/\sqrt{2n-1})$. Since $(V,N')$ is radially symmetric, $\mathbb P(0<N'<V/\sqrt{2n-1}) = \arctan(1/\sqrt{2n-1})/(2\pi) $. The probability of a sign change in each location is $\frac{1}{\pi} \arctan(1/\sqrt{2n-1}) = \frac{1}{\pi}\left((2n-1)^{-1/2} - \frac{1}{3}(2n-1)^{-3/2} + \frac{1}{5}(2n-1)^{-5/2} - \cdots \right) $. If you consider $n$ consecutive values for $Y$, there are $n-1$ possible sign changes, and the expected number of sign changes is $\frac{n-1}{\pi}\arctan(1/\sqrt{2n-1}) \sim \frac{\sqrt{n}}{\pi\sqrt{2}}$.