I am perplexed

$d_{1},d_{2},.....d_{2n}$ can be from $1,2,3,4,...,9$ only

$\sum_{j=1}^{n}d_{2j-1}d_{2j}$ is even If

$(1)$ all the $n$ terms are even

$(2)$ the number of odd terms in the sum is even

$d_{2j-1}d_{2j}$ is even if

$(1)$ both $d_{2j-1}d_{2j}$ are even . There are 16 choices for this

$(2)$ one of them is odd the other one is even ; there are $2(5)(4)=40$ choices for this $d_{2j-1}$ could be odd and $d_{2j}$ even or vice versa ).

Note $d_{2j-1}d_{2j}$ is odd for $25$ choices .

For the sum to be even , an even number of such products must be odd . Number of ways in which $k$ of the $n$ terms are odd and the rest are even is $\binom{n}{k}.25^{k}(56)^{n-k}$ .

Total sum is even if $k's$ are even .

Total number of ways

$P_{n}=\sum_{k=0}^{[\frac{n}{2}]}\binom{n}{2k}25^{2k}56^{n-2k}$

Here $[.]$ represents floor function.

Let $Q_{n}= \sum_{k=0}^{[\frac{n}{2}]}\binom{n}{2k+1}25^{2k+1}56^{n-2k-1}$

$P_{n}+Q_{n}=(56+25)^{n}$

$P_{n}-Q_{n}=(56-25)^{n}$

Hence $P_{n}=\frac{81^{n}+31^{n}}{2}$ Now for $n=7,P_{n}=11452152534536$

The $n$ terms in the sum are the products of couples of digits, from $1$ to $9$ . Given $d$ digits, the number of even products $E(d)$ and odd products $O(d)$ out of the $d^2$ couples are

$\displaystyle E(d)=d\bigg\lfloor \frac{d}{2} \bigg\rfloor + \bigg\lceil \frac{d}{2} \bigg\rceil \bigg\lfloor \frac{d}{2} \bigg\rfloor = \bigg\lfloor \frac{d}{2} \bigg\rfloor \bigg(d + \bigg\lceil \frac{d}{2} \bigg\rceil\bigg)$

$\displaystyle O(d)= \bigg\lceil \frac{d}{2} \bigg\rceil ^2$

This is easy to see putting the digits in a $d \times d$ matrix.

Hence, the formula in @Shivam Jadhav's solution becomes

$\displaystyle P_n = \sum_{k=0}^{\lfloor \frac{n}{2} \rfloor} \binom{n}{2k} E(d)^{n-2k}O(d)^{2k} = \sum_{k=0}^{\lfloor \frac{n}{2} \rfloor} \binom{n}{2k}\bigg[\bigg\lfloor \frac{d}{2} \bigg\rfloor \bigg(d + \bigg\lceil \frac{d}{2} \bigg\rceil\bigg)\bigg]^{n-2k} \bigg\lceil \frac{d}{2} \bigg\rceil ^{4k}$

for $n$ odd.