I thought it might be instructuve and of interest to present an approach to evaluating the integral $\int_{-\infty}^\infty e^{-x^2}\cos(nx)\,dx$ that does not rely on direct integration. To that end we proceed.
Let $f(y)$ be represented by
$$f(y)=\int_{-\infty}^\infty e^{-x^2}\cos(xy)\,dx \tag1$$
Differentiating $(1)$ under the integral reveals
$$f'(y)=-\int_{-\infty}^\infty xe^{-x^2}\sin(xy)\,dx\tag2$$
Integrating by parts the integral in $(2)$ with $u=-\sin(xy)$ and $v=-\frac12e^{-x^2}$, we obtain
$$\begin{align}
f'(y)&=-\frac12y\int_{-\infty}^\infty e^{-x^2}\cos(xy)\,dx\\\\\
&=-\frac12yf(y)\tag3
\end{align}$$
From $(3)$, we see that $f(y)$ satisfies the ODE $f'(y)+\frac12yf(y)=0$, subject to $f(0)=\sqrt\pi$. The solution to this ODE is trivial and is given by
$$f(y)=\sqrt\pi e^{-y^2/4}\tag4$$
Setting $y=n$ in $(4)$ yields
$$\int_{-\infty}^\infty e^{-x^2}\cos(nx)\,dx=\sqrt\pi e^{-n^2/4}$$
Letting $n\to \infty$, we find the coveted limit is $0$.
Because $(\cos(x)-\sin(x)) \leq 1-x$ over the range of integration, an upper bound for the integral is
$$\int_{0}^{1} (1-x)^n dx.$$
But,
$$\lim_{n \rightarrow \infty} \int_{0}^{1} (1-x)^n dx = \lim_{n \rightarrow \infty} \frac{1}{n+1} = 0.$$
Because the original integral is positive for all $n$, the limit must be zero by the Squeeze theorem.
Best Answer
There is another elementary way to get going. First, notice that: $$\cos x - \sin x = \sqrt{2}\sin(\pi/4 - x).$$ Therefore, your integral, which we denote by $I_n$ is equal to: $$I_n = n2^{\frac n2}\int_0^1\sin^n(\pi/4 - x)dx.$$ On the other hand, you can use integration by parts twice to show that:
$$\int\sin^nxdx = \frac{-\cos x \hspace{3pt} \sin^{n-1}x}{n} + \frac{(n-1)}{n} \int \sin^{n-2} x dx.$$
You can combine these two to obtain an explicit formula for $I_n.$
Update: This turned out to be interesting if we restrict ourselves to only elementary calculus. First, write: $$I_n = n2^{n/2}\int_0^{\tfrac{\pi}{4}}\sin^n\left(\frac{\pi}{4}-x\right)dx+n2^{n/2}\int_{\tfrac{\pi}{4}}^1\sin^n\left(\frac{\pi}{4}-x\right)dx = $$ $$ = n2^{n/2}\int_0^{\tfrac{\pi}{4}}\sin^n\left(x\right)dx+n2^{n/2}\int_{\tfrac{\pi}{4}-1}^0\sin^n\left(x\right)dx = A_n+B_n.$$ Now, using $\sin x\leq x:$ $$|B_n|\leq n2^{n/2}\int_{\tfrac{\pi}{4}-1}^0|\sin^n\left(x\right)|dx=n2^{n/2}\int^{1-\tfrac{\pi}{4}}_0\sin^n\left(x\right)dx\leq$$ $$\dfrac{n}{n+1}\left(1-\dfrac{\pi}{4}\right)\left(\sqrt{2} - \dfrac{\pi}{2\sqrt{2}}\right)^n\to 0.$$ Therefore, we simply need to study $A_n,$ which satisfies a rather simple recurrence relation (from integration by parts): $$A_n = \dfrac{2n-2}{n-2}A_{n-2} - 1,\,\, A_1 = \sqrt{2}-1,\,\, A_2 = \dfrac{\pi}{2}-1.$$
Numerical evidence suggests that $A_n$ is positive, increasing and converging to $1$ as desired. Moreover, odd indices are of the form $\sqrt{2}a_n-b_n$ and the even indices are of the form $\pi c_n -d_n,$ where $a_n,b_n,c_n,d_n$ are all positive integers. In fact, one can easily find a closed form for $a_n:$ $$a_n = \dfrac{2^{3n}(n!)^2}{(2n)!},$$ while the closed form for $b_n$ seems to be very difficult, if possible at all.
However, based on what I have tried, finishing using only elementary approach from here seems rather difficult. A naive method is to use induction to prove the monotonicity. But that amounts to showing the following estimate: $$A_n>\dfrac{n}{n+2}$$ or equivalently: $$\int^{\tfrac{\pi}{4}}_0\sin^n\left(x\right)dx>\dfrac{1}{2^{n/2}(n+2)} (\dagger)$$ But above seems tricky as using the naive inequality $\sin(x)\geq\dfrac{2\sqrt{2}}{\pi}x$ on $[0,\pi/4]$ proves to be a too crude. I tried strenghtening it further by using: $$\sin(x)\geq \dfrac{3}{\pi}x\cdot 1_{[0,\pi/6]}+\left(\dfrac{6(\sqrt{2}-1)}{\pi}x+\dfrac{3-2\sqrt{2}}{2}\right)\cdot 1_{[\pi/6, \pi/4]}$$ which was again too strong.
Lastly, one can technically find a explicit formula by doing the following: $$\dfrac{A_{2n+1}}{(2n+1)2^n\sqrt{2}} = -\int_0^{\tfrac{\pi}{4}}(\sin^2(x))^nd(\cos x) = \int_{\tfrac{1}{\sqrt{2}}}^1(1-t^2)^ndt = $$ $$\sum_{k=0}^n\binom{n}{k}(-1)^{n-k}\cdot\dfrac{1-2^{-k-\frac 12}}{2k+1}.$$ But again, this is not the nicest looking sum to manipulate.
Update: An elementary proof was found in this question and it was just a repeated application of integration by parts. Namely, \begin{align}\int_0^{1/\sqrt{2}}\frac{t^n\,dt}{(1-t^2)^{1/2}}&=\frac{1}{n+1}\left(\left.\frac{t^{n+1}}{(1-t^2)^{1/2}}\right|_0^{1/\sqrt{2}}-\int_0^{1/\sqrt{2}}\frac{t^{n+2}\,dt}{(1-t^2)^{3/2}}\right)\\&=\frac{1}{n+1}\left(2^{-n/2}-\frac{1}{n+3}\left.\frac{t^{n+3}}{(1-t^2)^{3/2}}\right|_0^{1/\sqrt{2}}+\ldots\right)\\&>\frac{2^{-n/2}}{n+1}\left(1-\frac{1}{n+3}\right)=2^{-n/2}\frac{n+2}{(n+2)^2-1}>\dfrac{2^{n/2}}{n+2}.\end{align} Thus, the proof is complete and should completely fall in the elementary calculus scope.