Unless I made a mistake, this is not true.

Take the random variable $X$ to be $M$ with probability $\epsilon$, $-M$ with probability $-\epsilon$, and distributed as $N(0, \frac{ 1- 2 \epsilon M^2}{ 1-2\epsilon} )$ with probability $1-2\epsilon$. Here we take $M$ very large and $\epsilon M^2$ somewhat small, for instance $\epsilon = M^{-3}$.

The total variation distance is $$\int_{-\infty}^{\infty} \max \left( \frac{ e^{ - x^2/2}}{\sqrt{2\pi}}- (1-2\epsilon)\frac{ e^{ - x^2 (1-2\epsilon) / (1-2 \epsilon M^2)} }{\sqrt{ 2\pi (1- 2 \epsilon m^2)/ (1-2\epsilon)}} , 0 \right) dx $$

since the spikes at $-M$ and $M$ don't contribute.

Now the distribution of $(X_1 + X_2)/\sqrt{2}$ is

- $N(0, \frac{ 1- 2 \epsilon M^2}{ 1-\epsilon} )$ with probability $(1-2 \epsilon)^2$
- $N(M/\sqrt{2}, \frac{ 1- 2 \epsilon M^2}{ 2 -2 \epsilon } )$, with probability $2 \epsilon (1-2\epsilon)$
- $N(-M/\sqrt{2}, \frac{ 1- 2 \epsilon M^2}{ 2 -2 \epsilon } )$
- $\sqrt{2} M$ with probability $\epsilon^2$
- $0$ with probaiblity $2\epsilon^2$
- $-\sqrt{2} M$ with probability $\epsilon^2$

The last three are peaks and don't contribute to the total variation distance. The two shifted Gaussians are supported around $M/\sqrt{2}$, with exponential error, and thus contribute an amount exponentially small in $M$ to the total variation distance. So the total variation distance is

The total variation distance is $$\int_{-\infty}^{\infty} \max \left( \frac{ e^{ - x^2/2}}{\sqrt{2\pi}}- (1-2\epsilon)^2\frac{ e^{ - x^2 (1-2\epsilon) / (1-2 \epsilon M^2)} }{\sqrt{ 2\pi (1- 2 \epsilon m^2)/ (1-2\epsilon)}} , 0 \right) dx $$

plus a term exponentially small in $M$. This integral is greater than the integral above by a factor linear in $\epsilon$, which has size $M^{-3}$, which is much larger than the exponentially small factor factor, so the total variation distance is larger than the previous case.